Yeah, I'll give that a go as long as the motor isn't smoked.
The only way to burn the winding of the motor would be to stall the shaft while the motor is connected to 12 V. The specification you posted for the motor indicates stall current of 3.90 A, which would produce total power loss of 12 x 3.9 = 46.8 W. That is quite enough to destroy the winding within fairly short period of time (< 20s in my guess). So far, all the failures I read about on this forum pointed to a very different scenario - one in which, over period of time, the contact resistance between the brushes and the commutator steadily grows, allowing less and less current to pass thru the winding. In the end, the current passing thru the winding is so low that it cannot spin the rotor, and the lock stops working. Limiting the current flowing thru the winding down to only a few milliamperes on the one hand makes the lock stop working, but on the other hand, it pretty much eliminates the possibility of burning the winding. Take, for instance my lock. I measured the contact resistance between the brushes and the commutator at about 1200 ohms, while the resistance of the winding itself is mere 3 ohms. 1200 ohms connected to 12 Volts will allow only 0.01 Amperes to flow, and total power dissipation of 12 V x 0.01 A = 0.12 W. However, since the winding has resistance of only 3 ohms, the voltage across it will only be 3 ohms x 0.01 A = 0.03 Volts, and the power dissipation on the winding will be 0.03 V x 0.01 A = 0.0003 W. Power dissipation this small cannot burn the winding.
Bottom line - when you take the motor apart and find the characteristic black streak on the commutator, rub it off and the motor will work normally again. Also, if you cannot find a fine-grit sandpaper, try the red rubber eraser from an ordinary #2 pencil, they add a very fine abrasive into the rubber compound, and it will work just like a very fine sandpaper.
oh well!