No, the baseplate will not move if the two motors are applying torques in such a way that the rotors remain at rest.
To analyse this problem, it is best to consider the forces on the three parts (base disk and two rotors) individually, using free body diagrams. There are two motors, each connecting a rotor to the base plate, and the effect that each motor has is that each motor applies a torque of magnitude $\tau$ and (arbitrary) direction of clockwise. It it important to note that both motors need to apply torques in the same direction if you want the wheels to the remain at rest.
Note that because the motor applies a torque ($\tau$ clockwise) on the rotor, by extension of Newton's 3rd Law to moments, the rotor will apply an equal and opposite torque ($\tau$ anticlockwise) on the motor. As the motor is fixed to the base plate, this means torques (both $\tau$ anticlockwise) are applied to the base plate.
The rotors will attempt to rotate clockwise, but will be unable to due to the contact forces each rotor exerts on the other rotor (this assumes that the rotors act like gear wheels, and that no slipping occurs between the two rotors.) So, rotor 1 exerts a contact force on rotor 2 (magnitude R, direction "down" in the diagram shown). By Newton's 3rd Law, rotor 2 must exert a contacts force on rotor 1 (magnitude R, direction "up"). Finally, there may be a force that the baseplate applies to the centre of the rotor (and vice versa, 3rd Law), but we need more information before we know which direction that force is in.
So, how strong is this contact force? How big is R? Well, first of all, let's make an assumption: let's assume that the rotors are at equilibrium (i.e. it will remain at rest). If the calculations do not have any contradictions, and if the base plate turns out to also being in equilibrium, then this assumption will turn out to be correct.
So, if we treat each rotor as it being at equilibrium, we can analyse it using static methods. For a rigid body to be in equilibrium, two things must be true:
1) The net force acting on the body must be zero.
2) The net moment (of force) acting on the body must be zero about any point.
We can use fact 2) to determine to value of R:
For either rotor, add up the moments about the point in the centre of the rotor (note there is a force that the baseplate applies to the centre of the rotor. If we analyse moments about that point, we don't need to worry about that force in this step, because the perpendicular distance to that force is zero). Let $r$ be the radius of each rotor:
Sum of moments about centre of rotor:
$$\tau - rR = 0$$
$$\therefore R = \tau / r$$
Good. Now, we have to deal with the force that the baseplate applies to the centre of the rotor. In order to satisfy rule 1) for each rotor, that force will need to be equal and opposite to the contact force. So, now we know the magnitude and direction of all the forces and torques applied to either rotor (see the diagram above).
Now, we need to see if the base plate obeys both conditions for equilibrium. From earlier, I said that each rotor applies a torque of magnitude $\tau$ and direction anticlockwise. Also, because the baseplate applied a force to each rotor, by 3rd Law, each rotor will apply an equal and opposite force to the baseplate. Therefore, we can see all the external forces and torques applied to the baseplate, shown in the diagram above.
We can immediately see that rule 1) is met, as the only two forces acting on the baseplate are indeed equal and opposite. How about rule 2)? To do so, let's add up moments about, say, the middle of the baseplate. Note the distance between the two points the rotor connect to on the baseplate is $2r$.
$$-\tau - \tau + R(2r)$$
Sub in the value for R:
$$-\tau - \tau + \tau/r \times (2r) = 0$$
Therefore, rule 2) is met. This means the baseplate is also in equilibrium. This mean that if the whole system is at rest, it will remain at rest even if the two rotors are running in the same direction. The baseplate will not move.
