How much can I overvolt a brushless motor?
In asking the question, you seem to have assumed the brushed or brushless design of the motor is not important. That is not true. A brushless motor requires an electronic controller. The effect on the motor in that case is as much determined by the controller design as it is by the motor itself. You have another answer that addresses that.

For a permanent magnet brushed motor, at the voltage level in question, the effect of doubling the voltage is mostly a question of the effect of doubling the speed. With no load connected to the motor, will probably be ok at twice the rated voltage and twice the rated speed. The current drawn by the motor is primarily determined by the load, not the resistance of the motor. The current is given by  I = (V - E) / R, where V is the supply voltage, E is the back EMF (proportional to speed) generated in the motor and R is the winding resistance. If there is no load, E is practically equal to V when the motor is running at full speed. It is only reduced by the slight load due to the friction of the bearings and commutator and the air drag on the moving parts of the motor. If there is a load, doubling the speed may increase the load dramatically thus increasing the current and increasing the power developed in the resistance of the winding. The load will particularly increase dramatically if there is something like a propellor connected to the motor.

Even without a load, it is possible that doubling the speed may increase the heating due to internal friction and air drag to the point that the motor fails. In an inexpensive motor, vibration could also be a factor.