Some further research has pointed out the answer as affirmative (though I will leave the question unanswered for now, as I'm still perplexed by Ando's rank restriction, which now seems unnecessary).
Theorem V.3.3 of [1]. Suppose the matrix map $f$ is induced by a scalar function on $I\subset \mathbb{R}$ applied to its input's eigenvalues. Then $$ \partial f(X; \Delta)=f^{[1]}(A)\circ H\,\,, $$$$ \partial f(X; \Delta)=f^{[1]}(X)\circ \Delta\,\,, $$ where $\circ$ is the Schur product in the eigenbasis of $X$ and $f^{[1]}$ is the first divided difference map with $f^{[1]}(A)=Uf^{[1]}(\Lambda)U^\dagger$$f^{[1]}(X)=Uf^{[1]}(\Lambda)U^\dagger$ given a diagonalization of $A=U\Lambda U^\dagger$$X=U\Lambda U^\dagger$ and the definition of $f^{[1]}$ for diagonal matrices,
$$ f^{[1]}(\Lambda)_{ij}=f^{[1]}(\lambda_i,\lambda_j)=\begin{cases} \frac{f(\lambda_i)-f(\lambda_j)}{\lambda_i-\lambda_j}&\lambda_i\neq \lambda_j\\ f'(\lambda)&\lambda_i=\lambda_j \end{cases}\,\,. $$
Exercise V.3.15. If $f\in\mathcal{C}^1(I)$ then it is matrix convex on $I$ iff for all $A,B\succeq0$ with spectra in $I$ $$ f(A)-f(B)\succeq f^{[1]}(B)\circ (A-B)\,\,. $$
This confirms that we have an equivalent characterization of smooth matrix convex functions as in the scalar case.
[1] Matrix Analysis, Bhatia 1997, https://link.springer.com/book/10.1007/978-1-4612-0653-8