EDIT: The geometrical point in space associated with the instantaneous axis of rotation does not accelerate, if the linear velocity $v$ of the center of the rolling disc is constant (same as the angular velocity being constant, as the disc has fixed radius). It simply follows the disc at velocity $v$, always tracking the point contacting the ground.
However, the fixed, physical point on the disc passing through the IAOR is accelerating, as the below answer explains.
As you zoom in closer and closer to the part of the circle that touches the ground, it looks flatter and flatter, and more and more like this intuitive picture:
Left side moving up,
and right side moving down.
(All points are accelerating upward.)
The part touching the ground is in the center. It is clear that as a point on the circle progresses from being one represented on the right side of this picture to being one represented on the left, it is mostly just accelerating upward, with purely upward acceleration at the center of this picture.
The point touching the ground is being pushed up by the ground$^1$, and hence the acceleration is straight up. This doesn't mean the entire disc accelerates off of the floor though, because the rest of the disc is being pulled down by gravity. Why does this point accelerate up, then, and instead of being balanced out by the rest of the disc pushing it down?
Well, I imagine that's because it takes time for the compression forces to propagate through the disc; but by that that time, the disc has already rolled on to another point. Hence the cycle continues.
$^1$ In this case, it is useful to think of Einstein's principle of equivalence: we should be able to reproduce the exact same physics if we accelerate a rocket ship, and watch a disc being accelerated upward on the floor. It is no surprise that the acceleration of the point contacting the rocket floor is straight upward, event when the disc is rolling at a constant velocity. (This also clarifies why the disc doesn't randomly accelerate upward any more than the ground pushes it up: it has inertia, and follows Newton's first law.) If we imagine the floor of the rocket is a bit tilted relative to its acceleration, you will reproduce the physics for a ball rolling down a hill.