I'm not entirely sure how the problem works out with that choice. My inclination is to think there would be two angular accelerations, the usual one pertaining to the pulley's rotation about its center of mass and one describing the motion of the center of mass of the pulley about point A. The latter one would be zero.
EDIT:
If you choose A as the origin, the angular momentum of the pulley is given by
$$L_A = Rmv_{\rm cm} + I_G\omega = (mR^2)\omega_{\rm cm} + I_{\rm cm}\omega$$ where ##\omega## is the usual angular velocity about the center of mass and ##\omega_{\rm cm}## describes the motion of the center of mass about A. That term vanishes because the pulley doesn't rotate about A, but if we keep it around a moment, differentiating would give
$$\tau_A = \frac{dL_A}{dt} = (mR^2)\alpha_{\rm cm} + I_G\alpha.$$ The fact that the pulley doesn't rotate about A tells us ##\alpha_{\rm cm} = 0## but it doesn't say anything about the usual rotation of the pulley about its axis.
If you calculate the torque on the pulley about A, taking into account the weight and the reaction force from the support, you should end up with the same net torque as before. So in the end, you get the same equation as when you use G as the origin.
