I will assume, though you didn't say, that the ground field is $\mathbb{R}$. (For all I know, the argument below might fail when the ground field is finite, etc.)
Yes, when $n>2$, it works for $m > {{n-1}\choose2}$, but not for $m = {{n-1}\choose2}$.
The reason is the following: Let $W$ be the span of the matrices $S_1,\ldots,S_m$. Your hypothesis is that, under conjugation by $U^{-1}$, $W$ gets carried into the skew-symmetric matrices, i.e., the Lie algebra of $\mathfrak{so}(n)$, or, equivalently, $W$ is a subspace of the Lie subalgebra $$ L = \{\ UxU^{-1}\ |\ x\in \mathfrak{so}(n)\ \}\subset \mathfrak{sl}(n), $$ which is, of course, isomorphic to $\mathfrak{so}(n)$. Let $K\subset L$ be the subalgebra generated by $W$. Because conjugation by $U$ is an automorphism of the Lie algebra $\mathfrak{sl}(n)$, your hypothesis implies that conjugation by $U^{-1}$ carries $K$ into a subalgebra of $\mathfrak{so}(n)$. Hence, if $K=L$, then your hypothesis implies that conjugation by $U^{-1}$ carries $L$ to $\mathfrak{so}(n)$. Since $L$ has dimension $n\choose 2$, your original argument for $m={n\choose2}$ finishes the job.
Now, when the ground field is $\mathbb{R}$, it's a fact that, except when $n=4$, $\mathfrak{so}(n)$ has no proper subalgebras of dimension greater than ${n-1}\choose2$, so, with the hypothesis $m>{{n-1}\choose2}$, one gets $K=L$, and the above argument finishes the proof. When $n=4$, one has to do a separate argument, because $\mathfrak{so}(4)$ does have proper subalgebras of dimension $4$, for example $\mathfrak{u}(2)$. However, this is the only case, and it is easily verified that, when $n=4$ and $K\simeq\mathfrak{u}(2)$, then you still get the desired result.
Meanwhile if $m={{n-1}\choose2}$, then we can take $W = \mathfrak{so}(n{-}1)$ and find many pairs of matrices $U$ and $V$ with determinant 1 that carry $W$ into $\mathfrak{so}(n)$ but that do not differ by an orthogonal matrix.
Added remark: If one changes the question slightly and asks what is the smallest $m$ for which there exist $n$-by-$n$ matrices $S_1,\ldots, S_m$ that can be simultaneously skew-symmetrized by conjugation and are such that any two unimodular matrices $U$ and $V$ that conjugate the $S_i$ into skew-symmetric matrices must differ by an orthogonal matrix, then the answer (for $n>2$) is $m=2$. This is because, for $n>2$, the Lie algebra $\mathfrak{so}(n)$ can be generated by any two generically chosen elements.