When you turn off the power to the motor, it will generate a voltage until it stops spinning. If you had some way of putting that power back onto the supply, you could get an increase of voltage. However, your circuit cannot use that power, so there isn't a problem. Even so, the battery would just absorb the power without the voltage rising much.
There are a couple of other things that can cause a problem in your circuit. Firstly, the inductance of the motor will be large, so the current in it cannot stop instantaneously. You need a path for that current. The body diode in Q2 will provide that path, but you could destroy Q1 if Q2 isn't fitted. 555
If you want to slow down the motor quickly, Q2 will short it out and will do that for you. You don't really need the, because a big capacitor feeding the gate of Q2 would turn it on for long enough, and it doesn't matter if Q2 turns off slowly, as the motor should have been stopped by then.
R6 needs to be much smaller than R1. When you release the trigger, R1 and R6 form a potential divider, and so the trigger input won't go below 1/2 the supply, so it won't trigger. There is no advantage of having R6 large, as it only wastes energy when the trigger is pulled, which is when a big motor is running anyhow.
You should also put a small resistor, maybe 470 ohm, in series with R3, otherwise, if you turn it to zero resistance, a lot of current will flow.
You won't get a higher motor speed than you can get connecting the motor directly to the battery, and you won't recover any motor energy without a seriously complicated circuit.