I don't have good news for you.
Below is the discharge rate for Durcell batteries done by some company. Look at the DC label; this is a Duracell coppertop battery. The complete test can be found here: link.
There was no µA test (it would take too long), but I guess the voltage will drop below 1.5 V after about 1-2% capacity discharge. There will also be battery self-discharge (very slow). My guess estimated time for 0.3 mA (300 µA) before dropping to 1.5 V will be somewhere between 50-150 hours. You can test this; it's not that long.
I would suggest to use different kind of batteries. Alkaline and a 3.0 V requirement is just the wrong battery for this application.
You can also consider different battery chemistry. If you use some lithium AA battery, for example Energizer L91 - they have much more energy "available" before voltage drop below 1.5 V, but be careful - they have also higher initial voltage (about 1.7 V) and they are expensive (in my country they cost 2-4x more than alkaline Energizer or Duracell).
Image from Energizer Ultimate Lithium L91 datasheet: link
However if you want to pull 0.3 mA (300 µA) from battery for two years (over 17000 hours) - you need more than 5 Ah (5000 mAh) before the voltage drops below 1.5 V. This is probably too much for any AA battery available on the market.
There are also nickel-zinc AA batteries, they have nominal voltage 1.65 V, but they have less capacity (50% less than Energizer L91).
Another idea might be three batteries with a low-dropout linear regulator. Three AA batteries in series will provide more than 3.0 V until they are completely empty, but a linear regulator may be necessary for some devices - three new alkalines may have 4.95 V initially.
A typical DMM has a very high (but not infinite) input impedance, typically ~10Mohm or bigger.
Now suppose you have very long leads. This will also have some finite resistance. Forming a voltage divider (I'm moving all of the resistance due to leads to above the multimeter. Mathematically this is equivalent to having two leads with 1/2 the length on each side):
\begin{equation}
V_{out} = \frac{R_{DMM}}{R_{DMM} + R_{leads}} V_{in}
\end{equation}
Computing the equivalent lead resistance for 1km 24AWG wire on each side, we get \$R_{leads} = 166 \Omega\$.
Then with a 10Mohm dmm resistance,
\begin{equation}
\frac{V_{out}}{V_{in}} = 0.9999834
\end{equation}
Or an error of 0.00166%. You'd be pretty hard pressed even measuring this error with most multimeters, and errors from other sources will swamp any errors due to the voltage drop in the leads.
There is slightly more error if you add in the battery's internal resistance, but still not significant.
Best Answer
Batteries have an output impedance and when a battery is not designed to provide high currents it has a high internal impedance.
That means there is basically a resistor in series with the output of the ideal voltage source that is the battery. So the more current you try to draw from it, the more voltage is dropped across the resistance and lost.