Very low power: Most power-efficient way of storing charge for hours
A capacitor may be the least lossy store BUT the capacitor size needed may surprise you.

Energy in a capacitor = \$ \frac{C V^2}{2} \$
Energy when a capacitor is discharged from \$V_1\$ to \$V_2\$ =
\$E = \frac{C(V_1^2 - V_2^2)}{2} \$
Rearranging \$ C = \frac{2E}{V_1^2 - V_2^2} \$
Plug in some made up figures to get a feel for magnitudes.:
Vmax = 5 V, Vmin = 3 V.
E = 10 µWh = 36000 µW seconds = 0.036 watt seconds = 0.036 joule.

\$ C = \frac{2E}{V_1^2 - V_2^2} = 2 \cdot 3.6 \cdot \frac{10^{-2}}{5^2 - 3^2} = 7.2 \cdot \frac{10^{-4}}{16} = 4500 \$ µF.
Ceramic and most plastic capacitors would be liable to be low enough loss. 
Any electrolytic capacitor is liable to be excessively lossy.
 4500 uF is an immensely large value for a non electrolytic capacitor.

Storage at higher voltage will decrease required capacitance by the square of the voltage. eg at 50 V you need \$ \approx \$ 47 µF or less 

However, I'd at least consider battery storage. It's possible that a small Lithium ion battery may do the job. If you want say 36 mJ in 1 second that's \$ 0.036 \cdot 3600 \$ W  \$\approx 130 \$ W !!!. That's FAR more than you need for the stated application.

E&amp;OE - going to sleep at keyboard (long day on 3 hours sleep) - I'll post this now and check figures later and look at battery implications. 

More anon ...