Does fibrewise vanishing of cohomology imply its vanishing (non-proper case)?

Question

Let $f: X\rightarrow S$ be a quasi-compact, quasi-separated, flat morphism of schemes, with $S$ locally Noetherian. Also, fix an integer $i\geq 1$.

Question 1: Does $H^i(X_s,\mathcal{O}_{X_s})=0$, for all $s\in S$, imply that $R^if_\ast\mathcal{O}_X=0$?

The answer is yes, if $f$ is also assumed to be proper (flat base change+theorem of formal functions+easy direct proof for $S$ Artinian local). In general, I would expect that there are some simple counterexamples but, to my surprise, I didn't find anything in the literature - probably I simply looked in the wrong places! I should also add that I would be particularly interested in counterexamples in characteristic zero, i.e. where $f$ is a morphism of schemes over a field of characteristic zero.

Here is a variant of question 1:

Question 2: Assume in addition that $f$ has a section $e: S\rightarrow X$. What is the answer to question 1 then?

The "raison d'être" for asking question 2 in addition is that a colleague explained to me how, granting existence of a section, the assumptions above ensure that $H^0(X_s,\mathcal{O}_{X_s})=\kappa(s)$, for all $s\in S$, implies that $f_\ast\mathcal{O}_X=\mathcal{O}_S$. So, at least in this very special case, the fibres of $f$ "remember" what the global sections of $f$ are.

Answer 1

Here is one simple counterexample:

Example. Let $X = \mathbf A^2 \setminus 0$ and $S = \mathbf A^1$, where $f \colon X \to S$ is the first coordinate projection. The fibres are affine, so have no higher coherent cohomology.

But if $V \subseteq S$ is affine open containing $0$, then $f^{-1}(V)$ has an affine open cover by $U_1 = V \times \mathbf G_m$ and $U_2 = (V \setminus 0) \times \mathbf A^1$. Then the function $\tfrac{1}{xy}$ on $U_{12}$ is not a sum of functions on $U_1$ and $U_2$, so it gives a nonzero class in $(R^1f_* \mathcal O_X)_0$. $\square$

(Note that this map has many sections.)