There are two sources of voltage to consider
Due to the motor turning the rotor through the permanent magnetic field, the motor will induce a voltage of the same polarity as it is driven with. This voltage limits the power the motor can deliver (and the current it can draw) at higher rotation rates. This voltage will not exceed Vcc unless the external power is applied to the motor shaft.
The effect you may want to be concerned about is any attempt to interrupt the instantaneous current through the motor. Assuming that the motor is not spinning at the maximum speed so that the generated voltage is less than the applied voltage, there will be current flowing through the motor.
For this purpose, at the instant you turn off the Darlington the motor is just an inductor.
You can't instantly turn off the current. Current will flow because the magnetic field which holds the energy is a charge that must be discharged. The question is what must the instantaneous voltage be for the current to continue. In your circuit, the current will flow and pull the switched side of the motor toward an increasingly negative potential. The voltage will go negative enough to make the "low" Darlington input voltage, when referenced to the emitter, to be high enough to keep the transistor turned on. This would be about negative 1.2V.
While this may not hurt the transistor, it will increase your turn-off time. If you want to keep the Darlington from turning on, you must prevent the emitter from going below about 1.2v negative.
A Schottky diode connected with the cathode to the emitter and anode to ground would do the job.
The reverse breakdown voltage of the Schottky diode must be greater than Vcc, plus a safety factor.