This answers the easiest question posed (1), and addresses part of the more general question (2).

(1) "If there are convex polyhedrons that *cannot* be folded from convex polygons,..."

Yes, there is an abundance of such polyhedra. For example, let $P$ be a
cube. To unfold it to the plane, one must form surface cuts that combinatorially constitute a spanning tree of the vertices, often called a *cut tree* $T$.
(Any vertex not on $T$ will retain its 3D structure, and so obstruct
planar development.) A tree has at least two leaves. Let $v$ be
a vertex of $P$ at a leaf of $T$. Then the corresponding unfolding
forms an interior reflex angle of $3\pi/2$ locally about the image of $v$.
This is a local nonconvexity.

(2) "how does one characterize convex polyhedrons (...) that can be folded out of convex polygons?"

Generalizing the argument above leads to a necessary condition:
A convex polyhedron $P$ needs to have
at least two vertices of curvature $\ge \pi$
(and so incident face angles $\le \pi$)
in order to serve as leaves for a cut tree $T$
that unfolds to a convex polygon.
The cube has all vertices with curvature $\pi/2$,
so there is no "home" for the leaves of $T$.

This is, however, not a complete characterization.

Mathematical Proceedings of the Cambridge Philosophical Society, vol. 78, no. 3, pp. 389-403. Cambridge University Press, 1975. $\endgroup$