Your FET will have an internal source-drain diode as a consequence of its semiconductor construction. According to your FET's datasheet, it can carry up to 34 A continuously with a 0.9..1.2 V drop across it.
The motor won't act like a resistor so I'm assuming that its normal running current is 32/12 = 2.66 A. Please can you verify that value and provide its stall (start-up) current, too.
If the FET in your circuit is the way up you've drawn it... (CORRECTED)
When powered normally, the FET's Gate-Source will be forward biased and the FET will be on. The FET's internal diode will not be forward biased enough to conduct and all FET current will flow Source-Drain. When the supply is reversed, the FET will be off as its Gate will be a D1 diode drop lower than its Drain. The FET's internal diode will be reverse-biased and so you will have your protection.
If the FET in your circuit is the other way up to what you've drawn...
When powered normally, the FET will conduct the motor current. When the supply is reversed, the internal diode will conduct and your motor will run backwards. So no reverse protection that way up.
Note that if your motor is a brushed DC motor, which your circuit suggests it is, it will generate EMI, and were you to run and power this through a diode, it would make the EMI much worse. That's something to bear in mind if EMC is important to an application. Putting diodes in supplies for higher-frequency circuits is a bad idea for EMI, despite the odd designer regarding them as 'magic one-way wires'. I've done a fair amount of EMC testing and this was showed this time and again :-(
ORIGINALLY, THIS REPLY WAS BASED AROUND AN INCORRECT AND WRONG STATEMENT ABOUT THE FET OPERATION AND HAS BEEN CORRECTED ABOVE...WHICH LEAVES IT WITHOUT A POINT TO MAKE. BUT I'VE LEFT IT HERE SO THE COMMENTS ON IT REMAIN. THE ORIGINAL STATEMENT WAS:
If the FET in your circuit is the way up you've drawn it...
When powered normally, the FET's internal diode will carry the full motor current and the FET will be reverse-biased. When the supply is reversed, the FET will be off as its Gate will be a D1 diode drop lower than its Drain.
This circuit is therefore no different to putting a big diode in the supply and doing away with M1/R1/D1/C1.