Is it possible to have a Schur decomposition of a matrix $A=URU^H$ so that the upper triangular matrix $R$ only has real non-negative numbers on the diagonal?
I realize the diagonal of $R$ is typically made up of the eigenvalues of $A$ which are not necessarily positive or real, but since $U$ isn't completely unique maybe there's a way to set it up?
Not sure how to prove or disprove this.
No. A simple counterexample is any negative-definite matrix $A$. In this case, we have
$$ \sum_k r_{kk} = \mathrm{tr}(R) = \mathrm{tr}(RU^HU) = \mathrm{tr}(URU^H) = \mathrm{tr}(A) = \sum_k\lambda_k < 0 $$
where $\lambda_k$ are the eigenvalues of $A$ which are negative by assumption and $r_{kk}$ are the diagonal elements of $R$.
The freedom in $U$ reflects the freedom in the choice of bases for the eigenspaces of $A$ and does not alter the fact that the diagonal of $R$ consists of the eigenvalues of $A$. See the proof of Schur's decomposition.
It is possible to write $A=URV^H$ where $U$ and $V$ are unitary and $R$ is upper triangular (in fact, diagonal) with non-negative real numbers on diagonal. This is the singular value decomposition.