Zero gravity from the sphere itself, yes. Put a black hole (or a planet, or a star, or a crouton, or...) somewhere, inside or outside of the sphere, and you'll feel the gravitational field from that, but the field from the sphere itself will be zero inside.
Consider any arbitrary chunk of the sphere. It subtends some patch of solid angle with respect to you. The gravitational pull from that chunk is opposed by the pull of a similarly-shaped chunk on the opposite side of you. Get this: the size (and therefore the mass) of each chunk will grow as the square of its distance from you. Its gravitational pull will fall off as the square of its distance from you. Those two factors cancel, thus the pull from each piece is equal and opposite. That zero-pull argument works for any arbitrary piece in any arbitrary direction. Integrate that zero over the entire sphere, and you get zero.
Thanks.
Doesn't Einstein's description of gravity as a curvature effect of space-time also have to be considered? Are you saying that the curvature of space-time is apparent all through the thickness of this sphere, but becomes completely flat everywhere inside the hollow sphere?
What if I said that the curvature is centered on the center of mass, which is the center point of the hollow sphere, therefore gravity effects would continue upon a falling body all the way to the very center? At which point the object would become "weightless" but would be under crushing pressure as every point not exactly at the very center would be under the maximum space-time curvature effects. A human would likely be squashed to a tiny little ball of goo, in other words.