The answer to Question 1 is yes, which is precisely Lemma 3.6 in the paper:
Although the lemma is stated and proved for $R^3$, the same proof works in $R^n$. The proof also indicates how to give a positive answer to Question 2.
The proof is elementary and proceeds as follows: a convex surface can be represented locally as the graph of a convex function over a convex domain in a support plane, which we may identify with $R^{n-1}$. Then the upper half of the sphere corresponds to the graph of a concave function over the same convex domain, assuming that the radius is sufficiently small and after we readjust the domain. Now the portion of the surface cut off by the sphere is the set of points where the convex graph lies below the concave graph. It is easy to check, using the standard inequalities for the convex and concave functions, that this portion projects onto a convex domain. Hence it is a disk.
In the setting of Question 2, we have a sequence of convex functions converging to the convex function mentioned in the previous paragraph. So eventually they will be defined over the same convex domain and lie below the concave function corresponding to the hemisphere, whence again we may conclude that the upper hemisphere cuts out a disk from their graphs.