In these cases it always helps to draw a diagram:
The green vectors represent the force of gravity $w=mg$ (dashed) and its components along the inclined plane and perpendicular to it. The red forces are the normal force of the plane on the ball $n$, the force of friction $F$, and their vector sum (dashed).
Now the sphere rotates about the contact point - that is the point that doesn't move. In that frame of reference, noting that the red vectors all pass through the center of rotation we compute the torque as the force of gravity $w$ times the perpendicular distance to the pivot point $d= r\sin\theta$, i.e. $$\Gamma = w\cdot r \sin\theta$$ and we consider the moment of inertia of the ball about this pivot to be $$I = \frac25 mr^2 + mr^2=\frac75 mr^2$$
(by the parallel axes theorem).
As you pointed out, by considering the motion about the contact point, the value of $F$ doesn't seem to come into play. But remember that the center of mass of the sphere must accelerate as though all forces are acting on it; after canceling out the normal forces, that leaves us with $mg\sin\theta$ down the slope, and $F$ going the other way. The difference between these two forces gives rise to the acceleration of the sphere's c.o.m. so we can compute $F$ from
$$mg \sin\theta - F = m a$$
To compute $a$, we first need the angular acceleration $\dot\omega$which is found from
$$\dot \omega = \frac{\Gamma}{I} = \frac{mgr\sin\theta}{\frac75 m r^2} = \frac{5g\sin\theta}{7r}$$
The linear acceleration $a$ is of course the angular acceleration multiplied by the radius of the sphere, so
$$a = \frac57 g\sin\theta$$
From which it follows that
$$F = \frac{2}{7} m g \sin \theta$$
And if we know that, we can now compute the angular acceleration of the sphere about its center. The torque seen in the frame of reference of the sphere is
$$\Gamma' = Fr = \frac{2}{7} m g r \sin\theta$$
Now we use the moment of inertia of the sphere about its center in order to compute the angular acceleration, and find
$$\dot \omega = \frac{\Gamma'}{\frac25 mr^2} \\
= \frac{\frac{2}{7} m g r \sin\theta}{\frac{2}{5} m r^2}\\
=\frac{5 g \sin\theta}{7 r}$$
which is the same result as before.
So there is no contradiction. The forces of friction and gravity work together to cause the rotation - the difference in apparent torque comes about from the fact that you are working in different (and non-inertial) frames of reference, but if you do the calculation carefully you get the same answer.
Now, acceleration, a = radius ( r ) x angular acceleration (α) Now as there is, a = gsinθ so there must be nonzero value of α.
That is only true for rotational motion. As it stands there is nothing in your problem to suggest that any sort of rotational motion will happen. Since the plane is "FRICTIONLESS" (your words), it does not do the normal "rolling without slipping" motion but instead slips completely, sliding down without any rotation whatsoever.
This is why, as you say, the net torque about the center will be zero and angular acceleration will be zero.
Angular momentum will may or may not be conserved at the bottom of the incline; it depends on whether the new ground that it is rolling onto is frictionless (hence torquing the object) or not.
Best Answer
That's actually a really interesting question. Stating it in a slighly different way:
And the answer is - no it would not. The reason for this is that the external force on the sphere (the normal force of the plane) acts through the center of the sphere. In other words - there is no "arm" and thus no torque.