Reachability queries on a tree in $O(1)$ time with $O(n+m)$ time preprocessing
Yes, I think you can do this in $O(1)$ time.  I outline a method below.  There's probably a simpler way to do it, but the following should suffice.

Preprocessing. I'll assume we can view your tree as a rooted tree, with some root.  In pre-processing, annotate each internal node $w$ with one example of a leaf that is a descendant of $w$.  This can be done in $O(n+m)$ pre-processing time, using a very simple top-down traversal (DFS).

Also, define $p^*(\cdot)$ recursively as follows: if $w$ has any siblings, then $p^*(w)=w$; otherwise, $p^*(w)=p^*(v)$ where $v$ is the parent of $w$; and $p^*(r)=r$, where $r$ is the root of $T$.  In English, $p^*(w)$ is defined to be the node obtained by starting at $w$ and continuing upwards until we reach a node that has two or more children (or the root).  In pre-processing, we'll compute the value of $p^*(w)$ for each node $w$ and store it associated with $w$.  This too can be computed in $O(n+m)$ time.

Answering a query. Now, here's how to answer the query return any leaf that is reachable from $v$ with a path including $(u,v)$, and no other edges incident to $v$:


Case 1: Suppose $u$ is a child of $v$.  Then, answering your query amounts to returning any leaf that is a descendant of $u$.  This can be done in $O(1)$ time using the annotation on $u$.
Case 2: Suppose $u$ is the parent of $v$, and $u$ has at least one other child.  Then, look at any other child of $u$, call it $w$.  Then, answering your query amounts to returning any leaf that is a descendant of $w$.  This can be done in $O(1)$ time using the annotation on $w$.
Case 3: Suppose $u$ is the parent of $v$, and $u$ has no other children.  Then let $u'=p^*(u)$ and let $w$ be any other child of $u'$ (not an ancestor of $v$).  Then, answering your query amounts to returning any leaf that is a descendant of $w$.  This can be done in $O(1)$ time using the annotation on $w$.


In all three cases, the answer to the query can be computed in $O(1)$ time.