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
As it is stated, the answer is "we don't know". It follows from the $O(\cdot)$ notation — roughly speaking, you don't have enough information to infer anything. You only know the asymptotic behavior, and not even the constants.
Algorithm 1 might very well be as fast as Algorithm 2 as long as $n$ is less than, say, $10^{10}$, and then break down.
Even worse, the big-Oh notation is only an upper bound. For all we known, both could have the same running time $O(n\log n)$, since if Algorithm 1 had running time $O(n\log n)$, it would also trivially have running time $O(n^2)$.