No, it fails even in the affine case. Let $R$ be a rank-1 valuation ring whose nonzero maximal ideal $m$ satisfies $m^2 = m$ (e.g., the valuation ring of an algebraic closure or completed algebraic closure of a complete discretely-valued field). Let $\pi \in m$ be a nonzero element of the maximal idea, so the fraction field $K$ of $R$ is equal to $R[1/\pi]$. Let $R' = K \times R/m$. Then the natural map $f: {\rm{Spec}}(R') \rightarrow {\rm{Spec}}(R)$ is a counterexample in the affine case because $R \rightarrow R/m$ is formally etale.
What underlies this is the fact that if $A$ is a ring and $J$ is an ideal such that $J^2 = J$ then $A \rightarrow A/J$ is formally etale (that being a non-flat map when $J \ne 0$, so perhaps slightly surprising at first sight, but not really surprising if one reflects on the logical structure of things). This is the standard counterexample to EGA 0$_{\rm{IV}}$ 19.10.3(i) and to EGA IV$_4$ 18.4.6(i) (whose "proof" ends by invoking EGA 0$_{\rm{IV}}$, 19.10.3(i)).