How long can a capacitor stay charged?
In case  of discharge through a resistance, there is an exponential law:
$$ V = V_{0} \cdot {e^{-t \over RC}} $$
but to get the time you need R and C: you have provided only V.

This figure shows it in a graphical way (for V=1V and RC~=1):



In your case, it will lose the charge creating a current over the lightbulb which will consume power (equal to VxI) to produce light for your eyes :) (and heat). The lower the bulb resistance, the greater the current supplied, and the shorter discharge time.

Note that the current will have the same curve of the voltage, as their ratio is given by the resistance.

This case is appliable also without load, if you know the equivalent resistance between the two plates (equivalent leakage resistance).

And this is only the basic transient, with R fixed. If you have a constant current, so it becomes a linear discharge with law:
$$ V={Q \over C} = {Q_0 - t \cdot I \over C} $$