Is there a way to relate convexity to Gaussian curvature?

For the final question posed:

Is it at least true that the boundary surface of a convex subset of $\mathbb{R}^3$ has non-negative curvature everywhere that its curvature is well-defined?

The answer is yes. Convexity implies that for every point $p$ on the boundary $\partial \Omega$, we can find a plane $\Pi$ through $p$ such that $\partial\Omega$ lies "on the one side" of $\Pi$. This implies that the second fundamental form is signed (either positive-semi-definite or negative-semi-definite). For surfaces in $\mathbb{R}^3$, the Gaussian curvature is the determinant of the second fundamental form, and hence is the product of the eigenvalues (the principal curvatures), and hence is always positive definite.

Now, as a general comment, convexity is more closely related to the extrinsic curvature of the boundary $\partial\Omega$ rather than the induced intrinsic curvature of it. It just happens that that for embeddings into Euclidean spaces (and especially into $\mathbb{R}^3$), we have easy relations between intrinsic and extrinsic curvatures; this drives what you observe with Gaussian curvature.

In fact, we have the following theorem:

Thm Let $\Omega\subseteq \mathbb{R}^n$ be an open domain, such that $\partial\Omega$ is an orientable smooth codimension 1 Riemannian submanifold. Then $\Omega$ is convex if and only if there exists a choice of orientation on $\partial\Omega$ such that the associated second fundamental form is positive semi-definite.

Sketch of proof:

(1) Essentially by Taylor expansion we see that "the second fundamental form is positive semi-definite everywhere" is equivalent to "locally $\partial\Omega$ is supported by a hyperplane". (The $\Leftarrow$ direction is obvious by taking local parametrization of $\partial\Omega$ as a graph over the hyperplane. As is the $\Rightarrow$ direction if the second fundamental form is positive definite. For the positive semi-definite case, when we encounter a zero eigenvalue, we need to use the fact that the form is positive semi-definite in an entire neighborhood to rule out the case of "bad" higher order Taylor coefficients.)

(2) Once we have the result of the first step, we can appeal to Tietze's theorem which establishes the equivalence between local convexity (existence of local supporting hyperplanes) and global convexity. See the link for a bit more discussion.

In the case of $n = 3$, as mentioned above the positive-semi-definiteness of the second fundamental form is directly tied to the non-negativity of the Gaussian curvature. For higher dimensions, the scalar curvature only controls the second symmetric polynomial of the eigenvalues of the second fundamental form, so is insufficient to control the local convexity of the boundary.