You can use any number of Dynamic Connectivity data structures, namely, an euler tour dynamic connectivity tree.
To summarize the reference, if you label the vertices of the original tree $T$ according to the first (denoted $label_f$) and last (denoted $label_l$) euler-tour visit order on $T$, you can maintain a BST $T^*$.
Each node $x$ in $T$ is linked to two nodes in $T^*$: $v_1$, $v_2$, where $key(v_1) = label_f(x)$ and $key(v_2) = label_l(x)$. Note $|T^*| = 2n $ and BST operations on $T^*$ will cost $O(\log n)$
Now, we can support the following operations on $T$:
- $cut(u, v)$: Disconnect the edge $(u,v)$. Denote $x$ the vertex whose $label_f$ is lesser. It is done simply by splitting from $T^*$ the interval $[label_f(x),label_l(x)]$ which can be done in $O(\log n)$ by splitting and merging.
- $findroot(v)$: Since the root of the tree has the minimum $label_f$, the root of $v$ will be the minimum node in the euler-tour BST in which he belongs
Since each node $x$ in $T$ is represented by exactly two nodes in $T^*$, by simply saving rank (tree size) on the euler tour BST ($T^*$), and dividing by $2$ you can answer the question "How many nodes are in the tree of $x$?" in $O(1)$, granted you are willing to pay $O(\log n)$ each time you disconnect an edge.
