The fact that the $H_i$ are matrices is irrelevant; the question is whether a module generated by elements $H_0\dots,H_n$ is necessarily generated by all the elements of the form $\Sigma_{0\le i\le n}r^iH_i$. If this is true in modules of matrices then it's true in all free modules, and if so then it's true when the $H_i$ form a basis for a free module, and if it's true in that case then it's true in all modules, in particular modules of matrices.
When the $H_i$ are a basis for $R^{n+1}$ then the question becomes "Is the ideal generated by all $(n+1)\times (n+1)$ Vandermonde determinants the unit ideal?" That fails if and only if there is a maximal ideal $m$ containing all such determininantsdeterminants, so if and only if there exists $m$ such that over the field $R/m$ you can make an there is no invertible Vandermonde matrix, so it fails if and only if some residue field of $R$ has at most $n$ elements.
The fact that the $H_i$ are matrices is irrelevant; the question is whether a module generated by elements $H_0\dots,H_n$ is necessarily generated by all the elements of the form $\Sigma_{0\le i\le n}r^iH_i$. If this is true in modules of matrices then it's true in all free modules, and if so then it's true when the $H_i$ form a basis for a free module, and if it's true in that case then it's true in all modules, in particular modules of matrices.
When the $H_i$ are a basis for $R^{n+1}$ then the question becomes "Is the ideal generated by all $(n+1)\times (n+1)$ Vandermonde determinants the unit ideal?" That fails if and only if there is a maximal ideal $m$ containing all such determininants, so if and only if there exists $m$ such that over the field $R/m$ you can make an invertible Vandermonde matrix, so it fails if and only if some residue field of $R$ has at most $n$ elements.