How to measure temporary drop in voltage with meter

Assumed the current you want to measure is indeed about 10uA and you feel OK with a burden voltage of about 1V and your multimeter has a voltage range of about 1V, then all you have to do is put a resistor is series with the FET you want to test. Most multimeters have a voltage range where 1V is somewhat in the middle of the range. In this case:

$$R_{series} = \dfrac{U_{burden}}{I_{expected}} = \dfrac{1\text{V}}{10\mu\text{A}}=100\text{k}\Omega$$

and measure the voltage across the resistor. The voltage measured is a pretty good indication of the current through the resistor, but it is important to check your meter's impedance. For good results it should be at least 10 times higher than the resistor. In this case the error will be better than 10% (\$\frac{100\text{k}\cdot 100\%}{1\text{M}}=10\%\$).

The leakage current will flow through the resistor and Ohm says:

$$I=\dfrac{U}{R}$$

So the voltage across the resistor is proportional with the current through it. R is known, U is measured, so you can calculate current I.

The voltage on a one farad cap will drop one volt per second given a uniform current draw of one ampere. Given three of the four values (capacitance C, acceptable voltage drop V, required hold time T, and current I) one can determine the value of the fourth using the equation CV=IT.

With regard to the capacitor and diode idea, you need to be aware that if one connects a large high-quality initially-discharged capacitor to an "already-on" power supply such as a battery, the cap will try to draw as much current as the supply will put out. It's conceivable a diode alone may work, but it would more likely turn into a "fryode" when the battery is connected; such a direct connection may also seriously damage the battery or the capacitor. Thus, rather than simply using a diode, you should use some kind of circuit which can safely handle that condition (charging a 1 farad cap to 12 volts will require 1200 amps for 10ms, 120 amps for 0.1 second, 12 amps for one second, or 1 amp for twelve seconds).

From an efficiency standpoint, you may be best off using a pair of switching-power-supply circuits, one of which would step your capacitor voltage up to some higher voltage, and one of which would step that higher voltage down to whatever your device needs. If your cap is charged to 12 volts, and your circuit will fail when it drops to 10, you'll only be able to use about 30% of your caps energy storage ability (the circuit will fail when the cap still has has about 70% of the energy that was put in). By contrast, if you were to charge a cap up to 35 volts and could operate down to 11 volts, and if you used a switching supply to extract energy from the cap, you could probably utilize more than 80% of the cap's energy storage capability. Given that large caps are expensive, the reduction in cap size may more than make up for the cost of the electronics.