Yes, because a tiebreaker can be modeled as changing the edge weights very slightly to make all edge weights unique.
Suppose that edges $e_1, e_2, \dots, e_m$ have weights $w_1, w_2, \dots, w_m \in \mathbb R$. Let $\epsilon$ be the smallest difference between two distinct edge weights. Suppose also that we've written edges $e_1, e_2, \dots, e_m$ in the tiebreaking order: whenever we are choosing between edges $e_i$ and $e_j$ with the same weight, we pick $e_i$ if $i<j$ and $e_j$ otherwise.
This decision process is equivalent to the decision process where edge $e_i$ has weight $w'_i = w_i + \frac{i}{2m} \cdot \epsilon$. The weights $w_1', w_2', \dots, w_m'$ are all unique, but comparing $w_i'$ and $w_j'$ is equivalent to comparing $w_i$ and $w_j$ when $w_i \ne w_j$. Only when $w_i = w_j$ does the extra term $\frac{i}{2m} \cdot \epsilon$ come into effect, and using it to compare is equivalent to using the tiebreaker.
So Kruskal with tiebreaks is equivalent to Kruskal with weights $w_1', \dots, w_m'$. This is equivalent to Prim with weights $w_1', \dots, w_m'$, because there is only one MST when weights are unique. That, in turn, is equivalent to Prim with the same tiebreaks.
Best Answer
You can use Kirchhoff's theorem. Let $D$ be the diagonal matrix whose entries are the degrees of the vertices of G. If you delete any row of $A-D$ (the Laplacian matrix) and its corresponding column, then the determinant of the resulting minor is the number of spanning trees for the graph.