but might 36v from a pair of panels damage the actuator circuitry?
So here's the deal. Lead-acid batteries look electrically like a voltage source/sink with a small series resistance, with the voltage level a function of state of charge. 2V/cell (there are 6 cells in series in a 12V battery) is nominal, and if I remember right, their open circuit voltage is something like 1.9V empty, 2.1V full. That covers 90% of their behavior.
Considering that, the "1W@18V" spec of the solar panel isn't going to be able to "win" against the battery, and the solar panel's voltage will be pulled down to battery voltage, delivering probably 0.055A (=1W/18V) at whatever the battery voltage is.
When a battery gets completely full, however, its series resistance goes up dramatically, and the voltage goes up, until there's enough voltage to start electrolysis of the fluid and you get H2 and O2 generation at the terminals and loss of the electrolyte. A lead-acid battery, depending on the type + manufacturer, has a certain recombination rate of H2 + O2 => electrolyte that it can handle; if you electrolyze at a higher current than that, it leads to permanent electrolyte loss (+hence capacity loss)
So there is a safe current that can be delivered to a lead-acid battery continuously, where its own self discharge due to electrolysis balances the charging current. It depends on the manufacture + construction. I wouldn't feel worried about a C/10 or C/20 rate of charge (where C = the current needed to discharge a battery in 1 hour). Garage door batteries are probably > 1Ah capacity so you should be safe with 55mA charging current.
HOWEVER -- I would probably put a (zener diode and resistor in series) in parallel with each battery, the zener diode being about 14V and resistor being maybe 10 ohms or so, so that it keeps the battery terminals from getting charged too far.
Also: if you can, wire each solar panel to each battery (and keep the diodes), rather than the pair of panels in series wired to the batteries in series -- i.e. try to connect the center taps. By doing so, you'll charge each battery independently. Otherwise, what can ruin battery life is if the battery voltages diverge -- the one with the higher voltage will tend to get overcharged, while the other one will tend to get overdischarged and not completely charged.
What is the frequency of the oscillation you observe?
If it is relatively low frequency, then the drop is probably caused just by energizing the TPS61222 output cap and inductor. You could fix that by adding more hysteresis on the comparator. Seems unlikely to me, however, because the super cap is so large.
If it is highish frequency, it could be caused by inductive and resistive drops in the path between the supercap and the input to the TPS61222 (including the internal resistance of the supercap itself, which is likely considerable). Again, adding more hysteresis on the comparator will help fix it. You can also improve the situation by using a bulk input capacitor right next to the inductor to hold the input to the TPS61222 in the face of the relatively large currents you will see at startup - this will work if you take Simon's recommendation and drive the EN pin of the boost converter instead of a load switch. That recommendation is likely a good one anyways, since it saves you a part.
Note that if you are running a load at the output of the boost, you may observe some sort of "oscillation" any time your load power exceeds the input power from the cell. The solar cell will charge up the super cap until the boost turns on, at which point the power drawn from the SC exceeds that going in and the voltage starts to drop. It continues dropping until the TPS61222 turns off / gets disconnected, and then the process starts over.
Best Answer
It probably won't damage them.
Any light absorbed by a solar cell will either be converted to heat or electricity. If the electricity generated is not drawn off then it will be absorbed by the intrinsic diode in the cell and converted to heat.
The worst case is with no load and one panel shaded on all cells but one. Because the shaded cells in that panel are producing a lower voltage, the non-shaded one gets a greater proportion of the voltage produced by the other panels, as well as having to absorb about double the current (the current it is producing plus the current pushed into that panel from the other panels). In a 6 cell panel that one cell will have to dissipate about 2.25 times more power from generated current than it would normally. Thus that cell will become a 'hot spot' in the panel.
Sounds bad, but the cell already dissipates ~75% of the incident light anyway, so total dissipation only increases by ~25% which won't heat it up much more. In a large panel it might be a problem, but small panels generally have a higher surface area to volume ratio so they can dissipate heat better.
Without blocking diodes any shaded panels will pull the voltage down a bit, but under load the effect is small. Diodes also drop voltage, so the loaded voltage will probably be lower with them than without. Suitable Schottky diodes (eg. 1N5820, rated for 3A max) could drop 0.3~0.4V at full current, which might be unacceptable on a system expecting 3.3V.