Why should a motorcyclist hit the brakes to make his motorcycle tilt in midair?

As you already noted, the center of mass should be the best point for analyzing the motion, because you don't have any external torque when you view the system from the center of mass. The general expression of angular momentum of any rigid body about any point is

$$\mathbf L=I_{\text{COM}}\boldsymbol{\omega}+m\mathbf r \times \mathbf v_{\text{COM}}$$

where $I_{\text{COM}}$ and $\mathbf v_{\text{COM}}$ are the moment of inertia of the rigid body (about the center of mass) and the velocity of the center of mass of the rigid body (when viewed from the point of reference) respectively.

In your case, the above formula can be applied to the wheel. Since the wheel is stationary (stationary in the sense of translatory motion) in the center of mass frame, the second term reduces to zero, and the total angular momentum of the wheel (about the center of mass) would then be

$$\mathbf L_{\text{wheel}}=I\boldsymbol{\omega}$$

Thus since the $\omega$ would be naturally high, the wheels do have substantial angular momentum about the center of mass.

Your professor is right. The angular momentum of the front wheel is measured relative to its spin axis. By touching the brakes, a little of that angular momentum is transferred via friction forces to the motorcycle and the rider. If the bike is in a nose-down state, a twist of throttle will bring the nose up because of the counter-torque that the engine applies to the motorcycle frame. But please don't try this at home!

As you have noted, angular momentum is always measured with respect to some point, but that can be any point. While there are some nice properties associated with the centre of mass, such as the fact that the principal axes of rotation pass through that point (i.e. objects rotate about their centre of mass), conservation of angular momentum still applies regardless of the point about which it is calculated (as long as you use the same point throughout). Therefore, you are free to choose whatever point makes calculations convenient, which in this case is probably going to be the centre of mass.