Effects of avoiding a snubber circuit?
You do have a snubber circuit: D1. It's a simple case of a diode snubber.

With D1 open, the energy stored in the motor's windings will have to go somewhere else. In this circuit, it will probably raise the voltage at the junction of the motor and Q1 until Q1's source-drain breakdown voltage is exceeded, and it begins to conduct in avalanche mode. For IRF744N, that looks like 55V. Of course, this puts a lot of stress on the MOSFET. It must be able to handle all the energy stored in the motor's inductance thermally. The datasheet also specifies a maximum repetitive avalanche energy of \$9.4mJ\$. If you can be sure the stored energy in the motor doesn't exceed that, and also that you don't exceed the thermal limits of the MOSFET, then you can operate without D1. However, given the complexity of calculating and designing for all of that, it's usually easier to just include D1. It's unlikely your motor is small enough to not exceed the MOSFET's limits anyhow.

Furthermore, if you are using PWM control here, the behavior of the motor will be very different without D1. The current in an inductor (motors being no exception) changes at a rate proportional to the voltage across them:

$$ v = L\frac{\mathrm{d}i}{\mathrm{d}t} $$

When Q1 is off, then \$v = 0.65V\$ with D1. The current will decrease slowly. Without D1, then \$v = 55V\$, and the current will decrease quickly. This means higher current ripple, and lower average current at a given duty cycle if D1 is open.