Let $f:X\to S$ be a finite type affine morphism of schemes where $S$ is an integral noetherian affine regular scheme whose function field is of characteristic zero.


Assume that all geometric fibers of $f$ are non-empty and that the generic fibre of $f$ is smooth.


Does have $f$ have a section up to replacing $S$ by some fppf covering?


The answer is positive if $f$ is flat, because then $f$ has a section after base-change along $f$.



Sections of morphisms up to fppf covering
The answer is negative; here is a counterexample. Let $S = \mathrm{Spec}(R)$ be the spectrum of a DVR whose fraction field of characteristic $0$ (to fulfill your requirements) and take $X := \eta \sqcup s$, where $\eta$ is the generic point and $s$ is the closed point of $S$. Then for any fppf cover $S' \rightarrow S$ we will have $X\_{S'} = S'\_{\eta} \sqcup S'\_s$, and hence $X\_{S'} \rightarrow S'$ will have no section since every open neighborhood of every point $s' \in S'$ lying above $s$ necessarily intersects $S'\_{\eta}$ (fppf morphisms are open).


