Interesting question - I would say not with a standard multimeter no.
One problem is (if it's similar to one I have seen) the voltage will be very high (>kV region) and the other is it is probably not stable DC, rather I think it is very short peaks (which will probably confuse even the AC setting on most multimeters) although further research says this is probably not the case.
In any case, the amount of charge supplied should be very small so it will probably be quite safe. Having a look at the circuit and running the calculations might be a good idea, as the voltage level (alone) is not the full story, the current matters, and although a capacitor can present a very low impedance, the output cap would need to be large enough to source it for a reasonable length of time. You might get a similar (harmless) shock from a piezo spark module found in e.g. a cheap lighter.
Wiki says:
Most flyswatters conform to electrical safety standards for humans:
a limit on the charge stored in the capacitor. A discharge of less than 45 µC is considered safe, even in the unlikely scenario that the current from a flyswatter would be flowing from one arm to the other arm, partly through the heart.1 This means that the capacitor of a 1000 V flyswatter should be less than 45 nF. Due to this precaution for humans the initial shock is usually inadequate to kill flies.
a limit on the current after the initial discharge. The maximal continuous current of most flyswatters is less than 5 mA. This current is safe, even when flowing from one arm to the other arm.[2]
To answer the question of voltage measurement:
If pulsed (or significant variation of output levels over small amount of time)
With the multimeter, if it short peaks you would need some way to average the voltage and reduce it to measurable levels - filtering with a capacitor and large restive divider would do this but the capacitors voltage rating (high) and leakage (low) would need to be suitable, plus it may be dangerous depending on the size of the capacitor (see above). You might need many megaohms of resistance in the divider to stop any "loading" of the voltage. The average current in needs to be higher than the current out so the capacitor can charge. I'm not sure if this method would be very easy to achieve satisfactory results with.
EDIT - If you can modify the circuit to stop the automatic discharge of the capacitor (i.e. the circuits output capacitor, not an external one as mentioned above) then measure with high resistance divider (e.g. 50:1, total resistance say, >10 megaohm), that is probably the safest way to try, though you will need to factor your multimeters input impedance into things.
An oscilloscope would be better, as there would be no need for the filtering and the input resistance is very high. You could maybe use a step down transformer to reduce the voltage and the impedance of the signal so loading was less of an issue, or use a HV (e.g. 100:1) probe to measure peaks.
If not pulsed (reasonably stable DC)
If it does not automatically discharge/spark or continually pulse charge the cap (e.g. charges up then holds), then measuring without modification using the divider mentioned should be possible, so would make things easier. I think this may be the case according to the Wiki and other pages (maybe the one I saw worked differently, I seem to remember it clicking/sparking)
Ok, after some additional research I think found some useful information from other manufacturers. I will just quote some points of interest.
From Technical Brochure LTC Batteries
Depending on mechanical cell design and system
properties, there is a certain dependence of available
capacity on cell orientation during discharge. The effect
is caused by the tendency of the electrolyte to move
towards the void and inactive space of the battery if the
orientation deviates from the preferred direction. The
capillary effect of the cathode and separator pores acts
against this tendency. As a result, the orientation effect
is smaller for thin cathodes than it is for thick ones and
is not even observable when discharge currents are very
low or when batteries are moved during discharge
[...]
At the high current end, available capacity of big cells
(C, D, DD) is affected if the batteries are discharged
upside down. Therefore this orientation should be avoided if possible.
From Handbook Primary Lithium Cylindrical Cells - VARTA
Under upside down installation, the capacity of smaller
size (1/2 AA, AA) is less affected whether discharge current is high, nominal or low. However, the capacity of bigger size (C, D) especially at higher discharge current is
affected. Under upside down installation, the lithium and
cathode is located in a fixed area whereas the electrolyte
falls to the bottom in this case. At the top of the cell there
is a space leaving an area of the anode and cathode, not
covered by the electrolyte. Bigger size cells have a bigger empty space, so the capacity decrease in upside
down installation is higher than in cells of smaller size.
(About 20~40% of its capacity at same higher discharge
current.)
While we are not "at the high current end" and are using a AA cell, the measurements from above confirm that "upside down" is the worst orientation.
So orientation does absolutely matter when using Lithium-Thionyl-Chloride cells. Learning never stops...
Best Answer
You are right in saying that the voltage across the cell gives an indication on its charge status, but the problem is that that indication is relative to the specific model of the battery. For example, two AA cells both having, say, 1.35V don't need to have the same amount of energy stored. That depends on many variables related to the specific chemistry used by their manufacturer.
Therefore, although you won't damage two batteries with the same voltage reading, they will discharge at different rates, probably. After a while one of the two will have a much lower voltage and a correspondingly higher internal resistance and will act as an additional load for the other.
Probably the only safe approach is to use cells of the same model from the same manufacturer and built almost in the same period: manufacturers improve their chemistries from time to time and I won't trust two batteries to be identical if one has been built two years after the other.
As for the range of the difference in voltage, 1%-2% is probably safe, i.e. from a practical POV if their voltage differ only by a couple of digits in the third digits you are ok: 1.25 and 1.23 is OK, 1.25 and 1.29 is borderline, 1.25 and 1.35 is NOT ok.
Anyway, I recommend to do this mix only with low power gadgets, in order to reduce risks to abuse a cell. That is, if you draw very little current, that's ok. If you try to use that mix-up in a power-hungry device it could overstress the weakest cell in the set and cause leakage or something worse.
EDIT (prompted by a comment)
The source for what I said above is just personal experience. After all, you are asking something that is contrary to every industry best practice, thus I interpreted yours as an hobbyists' question and answered in that context. If you asked me for a professional advice about that I'd replied not to ever mix and match cells like that. No one in a professional context would do that.
Don't get me wrong, I also reuse old cells, both because of environmental concerns and to avoid wasting usable energy, but I rarely mix two old cells "having a different history". When I change batteries from an "high-power" gadget I measure the old ones, if they are still usable, say V>~1.1V, I put them away sticking them together with a rubber band, so that they don't mix with other old cells. Then, if I have a low-power gadget (wall clock, small LED flashlight, digital thermometer, etc.) that needs that number of cells I'm sure they have all the same level of charge.
In rare cases I have done the mix and match, measuring their voltage as I told you before. What's the rationale of my 1%-2% rule of thumb? If the difference in voltage is so low, I can safely assume (being the same model and coming from the same batch) their internal resistance is different roughly by a similar percentage. That means that the load unbalance when used in a gadget in percentage will be roughly of the same order of magnitude (a few percent). Assuming I use them in low-power gadgets, i.e. in devices that draw say 5%-10% of the cell's max current, the cell with higher resistance will dissipate a few percent more than the already low power average dissipation of each cell, which is well below the safe threshold, since I'm working with loads that draw much less than the max current a cell can stand.