You need four pieces of information
1) How much current do you need (in amps)?
2) How long do you want to run your circuit from the cap (in seconds)?
3) What it the largest voltage (normally the regular operating voltage) which will be applied? and
4) What is the lowest voltage out of the capacitor which will still allow the circuit to function?
Let's call 1) i, let's call 2) $\delta$t, 3) is Vmax, and 3) minus 4) is $\delta$V.
The relationship between time, voltage and current in a capacitor with a value of C in Farads is $${\frac{i}{C} = \frac{dV}{dt}}$$
For a constant i, which is a reasonable first approximation in this case, this becomes $${\frac{i}{C} = \frac{{\Delta}V}{{\Delta}t}}$$, or$${C = \frac{i{\Delta}t}{{\Delta}V}}$$ and the capacitor must have a voltage rating of Vmax.
To walk you through this, lets say you need .25 amps for 5 minutes (600 seconds). Let's say your battery puts out 3.7 volts when fresh, and your circuit will work down to a battery voltage of 3.0 volts. Then $$C = {\frac{(.25)(600)}{(3.7 - 3.0)}}$$ $$C = {\frac{150}{0.7}}$$
and you need a 214 Farad supercap with a voltage rating of 3.7 volts minimum.
EDIT - As Brian Drummond has pointed out, supercaps often have a high internal resistance. If you want to take this into consideration (and you had better do so), you need to quantify the resistance R of the cap at the current level which you are using. Then the capacity calculation remains the same, but if the capacitor voltage rating is Vcap, instead of being Vmax, $$Vcap = Vmax + (iR)$$ In the above example, if the ESR of the supercap is 20 ohms, $$Vcap = 3.7 + (.25)(20)$$ and $$Vcap = 8.7$$ In this case you would definitely need to find a different supercap.
Yes, you can charge the two cells as a series 2.4V pack. During charging the voltage will rise to ~3V.
The charger should be current limited, to avoid charging the battery too fast and overheating it. The circuit could simply be a resistor in series which limits current to a safe 'trickle' level (10 hour rate = 100mA for a 1000mAh battery). The resistor might be inside the charger (perhaps explaining why the voltage drops when you try to draw power from it) or built into the iron - then the 'charger' is just a DC power supply with high enough voltage to make up for loss in the resistor.
While you have the iron disassembled you could look for a current limiting circuit. If it doesn't have one (ie. the battery is connected directly to the charge socket) then you must use a charger which is current-limited. Don't try to use a DC power supply, even if its ratings appear to be the same.
Most USB ports will deliver 100mA without any negotiation, so to charge from USB you just need a resistor in series which limits current to <=100mA. Assuming the cells are 1.1V when flat, the resistor must drop 5V-2.2V = 2.8V. Calculating the value using Ohm's Law gives 2.8V/0.1A = 28Ω (the nearest preferred value of 27Ω should be close enough). The resistor will dissipate up to 2.8V*0.1A = 0.28 Watts, so it should be rated at 0.5W or higher. As the battery charges up the voltage difference will reduce causing charge current to drop, so to get a full charge you may have to leave it on for 12~14 hours.
The original Nicad battery probably had lower internal resistance and could hold a slightly higher voltage under load, so you may find the iron takes a bit longer to heat up. However NiMH has higher capacity than Nicad per volume, so you might considering using similar size NiMH cells that have higher capacity - rather than smaller cells which have higher internal resistance and may not last as long. Avoid ultra-high capacity AA cells, as these are optimized for capacity rather than power.
Best Answer
Lets assume your trimmer draws 600 mA while running thereby giving you a backup of 1 hour on full charge.
Resistance of trimmer = 1.2/0.6 = 2 ohms
Now lets assume the power is being provided by a supercap. Lets further assume that your trimmer will work in the voltage range of 1.4 V to 1.1 V without any issues and you would need a backup of at least 20 minutes so that you can shave after one charge.
Refer to the discharge formula given below:
Vc = 1 V
Vo = 1.4 V
R = 2 ohms
t = 20 mins = 1200 seconds
Calculation gives C = 2000 F.
Yes, that's farads.
So, lets talk about feasibility. Here is a 2000 F capacitor from digikey:
This has a dia of 6 cm and height of 10 cm and costs 56 USD.
If this size and price point is okay with you, you can go ahead :)
Edit : Corrected capacitor specs.