It is convenient to consider the vertex set of $B_h$ as a partially ordered set with its natural genealogic order, with minimum element its root. Any subtree $T'$ as described in your procedure (also including the empty tree) is then exactly an *initial segment*. (For the notation: $B_1$ is the one-vertex tree, and so on)

Removing $T'$ from $B_h$ leaves a set of $m$ complete trees, whose $m$ roots constitue exactly an $m$-*antichain* in the partial order. The only $0$-antichain is the empty set and produces the empty forest. The $1$-antichains are the vertices as singletons; there are $2^{h}-1$ of them, including the root. The number of forests with *exactly* $m$ trees in $B_h$ obtained removing a $T'$ is therefore the number $a_{h,m}$ of all $m$-antichains in $B_h$. So in your question $|F_{h,m}|=\sum_{2\le j\le m}a_{h,j}$, since you like to consider forests with more than one, and no more than $m$ trees. There is a clear convolution formation passing from antichains in $B_h$ to antichains in $B_{h+1}$, giving the recursive relation for the generating polynomial $P_h(x):=\sum_{m\ge0}a_{h,m}x^m$, $$\cases{P_{0}(x)=1\\P_{h+1}(x)=x+P_h(x)^2.}$$

(and of course $P_{h}(x)/(1-x)$ counts antichains with at most $m$ elements in $B_h$). The problem of counting antichains in posets is well-studied. Here is a 2006 paper "*Counting chains and antichains in the complete binary tree*"
you may want to request to the authors (Though I'm not sure if it address the $m$-element version). Incidentally, note that $P_h(1)$, the number of all antichains in $B_h$, is the sequence A003095.

**Edit.** In fact, if you want an asymptotics for $a_{h,m}$ as $h\to+\infty$ for each fixed $m$, this is easily available: It follows by induction that for any $m$
$$a_{h,m} \sim {2^{mh}\over m!} \quad\text{ as }\;h\to+\infty, $$
so that the bound given by Emil Jeřábek in comment below in terms of $m$-elements subsets of $B_h$ is also an asymptotic, since
$$ {2^{mh}\over m!} \sim { 2^{h}-1\choose m } \quad\text{as }h\to+\infty. $$ Indeed we have, for $m=1$:
$$a_{h,1} =2^{h}-1\sim 2^{h}\quad\text{as }h\to+\infty,$$
while for $m>1$,
$$a_{h+1,m}=\sum_{j=0}^m a_{h,j}a_{h,m-j}= 2a_{h,m}+\sum_{j=1}^{m-1} a_{h,j}a_{h,m-j}$$
so assuming by complete induction hypothesis that $a_{h,j}\sim {2^{jh}\over j!}$ as $h\to+\infty$ holds for each $ j<m$, we find
$$a_{h+1,m}- 2a_{h,m}\sim {2^{mh}\over m!}\, \sum_{j=1}^{m-1} {m!\over j!(m-j)!} \,=\,{ 2^m-2\over m!} \,2^{mh}$$
or, dividing by $2^ {h+1 }$
$${a_{h+1,m}\over2^{ h+1 }}- {a_{h ,m}\over2^{ h }}\sim { 2^{m-1}-1\over m!}\,2^{(m-1)h} $$
whence, summing from $0$ to $h-1$
$${a_{h ,m}\over 2^{ h }}\sim {2^{m-1}-1\over m!}\,\sum_{j=0}^{h-1}2^{(m-1)j}={2^{(m-1)h}-1\over m!}\sim{2^{(m-1)h}\over m!} $$
so that
$$a_{h,m} \sim {2^{mh}\over m!} \quad\text{as }h\to+\infty.$$

any$T\in B_h$, $T'$ can beanysubtreeof $T$ such thatthe root of $T'$ is the same root of $T$, and(i)the number of subtrees of $T$ that we obtain by removing all nodes of $T'$ from $T$ (together with all incident edges in $T$), is equal to $m$. $\endgroup$(ii)19more comments