Limited number of calls to a black box for finding longest path in graph given an access to oracle
Since you can find the actual length of the longest path using binary search, in $\log |V|$ steps, I'll assume you have access to an oracle which tells you the length of the longest path. Given a graph $G$ lets denote the length of the longest path in $G$ by $L(G)$.

If you remove an edge after finding out that it does not affect the length of the longest path in $G$, then you avoid the issue of having to commit to some specific longest path. Suppose you keep iterating over the remaining edges, and for each remainning edge $(u,v)$ you check if
$L(G)=L\left(G\setminus\left\{(u,v)\right\}\right)$. If they are equal, remove $(u,v)$ from $E$ and return to the main loop. When there are no more edges to delete, $E$ consists of a longest path in $G$.

To prove this, let $\mathcal{P}$ be the set of longest paths in $G$. Note that in any iteration $i$ (each iteration consists of removing an edge which does not affect the longest path length), there exists some $P\in\mathcal{P}$ such that $P$ is a path in $G_i$. Using this observation, you can show that when there are no more edges to delete, your graph is a longest path.