Voltage

The use of your equation V = E/Q (where V is the voltage of the battery, E is the energy given out by the battery for an amount of charge Q) is irrelevant to the explanation for the drop in voltage of a battery.

The terminal voltage of a battery drops is a result of the increase of its internal resistance.

Battery rpoduces current by chemical reactions. This chemical reaction produced substances that accumulate at the electrodes of the battery. The accumulation of these substances would reduce the contact areas of the electroyle with the electrodes, casusing an increase of electrical resistance.

From Ohms Law, the voltage of a battery is given by
V = emf - I.r
where emf is the electromotive force of the battery (which is constant)
I is the current delivered by the battery
r is the internal resistane of the battery.

An increase of r will lead to a decrease of V

Now, using your equation, a drop of battery voltage will certainly give out less energy to an external circuit for the same amount of charge Q.