Is it true that the surface area of a convex hull of a non-convex polyhedron is not greater than the surface area of the non-convex polygon?
I believe this to be false. I believe that they can have equal surface areas. I picture a box with the corner pushed in. If we push the corner back out to form the box correctly, they would have the same surface area. I don't know how to explain this mathematically.
No. The surface area of convex hull can be much larger. Think of a dumbbell: namely, take two balls of radius $1$ with centers at $(\pm n,0)$ and connect them by an extremely thin cylinder. The resulting shape has surface area just a little over $2\cdot 4\pi = 8\pi$. But the boundary of its convex hull contains a cylinder of radius $1$ and height $2n$. So, its area is at least $4\pi n$. And $n$ could be arbitrarily large.
The above are not polyhedra, but you can just as well replace balls with cubes and connect them by a very thin squarish cylinder. The convex hull is a long rectangular box whose surface area is much larger than the surface area of the original polyhedron.
In the plane, taking convex hull does not increase the perimeter. But this does not generalize to higher dimensions.