Let $\pi:X\rightarrow S$ be a smooth scheme over $S$ and let's assume for simplicity that $S=\mathrm{Spec} R$ is affine, Noetherian and regular. Let $\mathcal F$ be a coherent sheaf on $X$ and let's assume as well that $\mathcal F$ is torsion-free. Then I can take the derived global sections $R^\bullet\pi_* \mathcal F$, taking further its cohomology I get some $R$-modules $R^i\pi_*\mathcal F$.

Now the question is:

Given $R^i\pi_* \mathcal F$ can I say anything about torsion in it (as in an $R$-module)? In particular is there an open dense $U\subset S$ such that $R^i\pi_*\mathcal F$ is torsion-free restricted to $U$?

If $\pi$ is proper this is automatic since then $R^i\pi_* \mathcal F$ is finitely generated over $R$. If $\pi$ is affine then $R^\bullet\pi_* \mathcal F=\pi_*\mathcal F$ and we just consider some torsion-free module $M$ over $R'$ as a module over $R$ under an injection $R\subset R'$, so the statement holds as well. For the general case I tried to use a Cech cover of $X$ by affine $S$-schemes, but what you get is a complex whose terms are infinitely generated $R$-modules (though torsion-free) and I'm not sure what I could use to bound the torsion in the cohomology there.

Also, not sure if this makes life any simplier, but under the assumptions above $\mathcal F$ is a perfect complex on $X$ and it would be enough to take $\mathcal F=\mathcal O_X$.