It is not $O(1)$ for sure. Take isomorphic (say, to $K_m$) disjoint graphs $G_1$, $G_2$, $\dots$, $G_N$ and draw one edge between $G_i$ and $G_{i+1}$, $i=1,2,\dots,N$. Then frequencies of edges of $G_i$ are the same and depend only on $m$, so each $G_i$ is chosen as whole with some probability depending only in $m$, These events are independent, so for large $N$ with high probability at least one of them holds. But for covering $K_m$ by trees you need at least $m/2$ trees.
And it looks probbale that it is $O(\log n)$ or even $O(\sqrt{\log n})$ or smth like this. Indeed, what does it mean that we can not cover graph by $K$ trees? That for some set $A$ of some $m$ vertices the graph has at least $K(m-1)$ edges between vertices of $A$. Fix $A$ and estimate the probability of this unlucky event. We have sum of independent variavles, and some of their expectations is nothing but the average number of edges covered by random spanning tree in $A$. So, at most $m-1$ for sure. Even if it we increase some probabilities so that it becomes $m-1$, dispersion is about $m$ too, so the probability of our event is estimated by some $\exp(-cK^2m)$ for some $c>0$ (I am not sure in this point, but this must be very standard weak version of CLT). If $e^{K^2}\gg n$, then even if multiply this by number of $m$-subsets of $V$, and then sum up by $n$, we get some infinitesimal.