It's easy to see that the compact-open topology agrees with the strong operator topology on norm-bounded subsets of $B(H)$. Bill Johnson mentioned this in a comment. I think this shows that of all the "usual" topologies on $B(H)$ the only candidates for agreeing with the compact-open topology are the strong and the ultrastrong toplogies.

However, I believe compact-open is strictly stronger than ultrastrong (which is itself strictly stronger than strong). To see this, fix an orthonormal basis $(e_n)$ of $H$ and consider the compact set $K = \{0\} \cup\{n^{-1/2}e_n: n = 1, 2, \ldots\} \subset H$. Then the set $U$ of all operators in $B(H)$ which take $K$ into the open unit ball of $H$ is open for the compact-open topology, but it is easily seen to not be strongly open --- given any operator $A$ in $U$ and any finite list of vectors $v_1, \ldots, v_k \in H$ you can easily find $B \in B(H)$ such that $Bv_i = Av_i$ for $1 \leq i \leq k$ but $\|Be_n\| > n^{1/2}$ for some sufficiently large $n$. (For large $n$ the vector $e_n$ is almost orthogonal to ${\rm span}(v_1, \ldots, v_k)$, so we have freedom in defining $Be_n$.)

It seems to me that a similar argument shows that the set $U$ is not even ultrastrongly open. Given any finitely many positive operators $C_i$ in the predual of $B(H)$, since their eigenvalues are square-summable, as $n$ goes to infinity we're going to have ${\rm max}_i \langle C_i e_n,e_n\rangle = o(n^{-1/2})$, so that again you can find $B$ which approximates the behavior of $A$ when tested against each $C_i$ but has $\|Be_n\| > n^{1/2}$ for some large $n$. That's just a sketch but I think the idea is sound.