The current taken from the capacitor will only be that current needed by the load. It's ohms law. If the capacitor is charged to (say) 3 volts and the load normally takes 1 mA at 3 volts (3000 ohms) then it's the same when connecting that 3000 ohm load across a charged capacitor. The capacitor doesn't magically emit several amps the second something approaches it. Regard it as a battery with a very defined voltage discharge curve.
Consider also that you might get more life by using a voltage regulator and charging a significantly higher voltage onto the capacitor that feeds the voltage regulator. Use a low power buck regulator for even more life between charges. Do the math!
Maybe this buck regulator: -
It will work with an input voltage as low as 2.5 volts and up to 15 volts and introduce a power loss of about 0.33 mW. Given that the load is about about 1 mW, the losses do seem quite high but, if the capacitor is charged to say 12 volt I think there are significant gains to be made.
If you have a 1 farad supercap charged to 12 volts you can work out the discharge time based on Q = CV.
\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$ and this equals current.
So, with an overall current of say 1 mA, dv/dt = 0.001 (capacitance is 1 farad). Voltage discharges at a rate of about 1 mV per second. In an hour, the 12 volts will have dropped by 3,600 mV to 8.4 volts. After two hours (ignoring the fact that current will inevitably reduce towards more like 0.75 mA), the voltage will be 4.8 volts. A further 30 minutes and the terminal voltage is 3 volts.
So, conservatively, you should get a shade over 2.5 hours with a 1 farad supercap charged to 12 volts.
Given the low current requirement it should also work with a linear LDO regulator like the LT1761-2 (2 volt version): -
There is a 2 volt fixed output version that will work down to an input voltage of 2.5 volts. The quiescent current of the device is 20 uA so this actually looks ideal for what you want. Maximum input voltage is 20 volts but, if we used the 12 volt limit as used previously, the overall discharge current is 520 uA and not the assumed 1 mA. This nearly doubles the discharge time from a 1 farad supercap charged to 12 volts.
However, finding a 12 volt supercap is going to be problematic and one rated at 5.5 volts seems more likely. 1 farad charged to 5 volts being discharged with 0.52 mA will lower its terminal voltage by 0.52 mV every second. So, in one hour the total reduction will be 1.872 volts taking the 5 volts down to about 3.1 volts.
You would probably achieve 1.5 hours using a low power linear LDO voltage regulator (like the LT1761-2) and a 5.5 volt 1 farad supercap.