Projective after fpqc base change
Hironaka's example is a locally projective birational morphism $f \colon \tilde X \to X = \mathbf P^3$ where $\tilde X$ is not projective; in particular $f$ is not projective. This already shows that the property of being projective is not Zariski-local on the target, so in particular not fppf-local.
But you want the base to be affine, so let's do a little bit more.
Lemma. Let $U \subseteq \mathbf P^3$ be any open containing the intersection $\{p,q\} = C \cap D$, and let $\tilde U = f^{-1}(U)$. Then $f|_{\tilde U} \colon \tilde U \to U$ is not projective.
In other words, the non-projectivity of $f$ is "purely concentrated above $p$ and $q$", and is not a global phenomenon. The idea is that away from $p$ and $q$ you can concretely write down what the $f$-ample line bundles are.
Proof. Suppose it were, and $\mathscr L$ on $\tilde U$ is a relatively ample line bundle. Let $Z = X\setminus U$. By the sequence \begin{align*} \mathbf Z &\to \operatorname{Pic}(\tilde X) \to \operatorname{Pic}(\tilde U) \to 0 \\ 1 &\mapsto [f^{-1}(Z)]\\ \end{align*} we can choose an extension of $\mathscr L$ to $\tilde X$. We claim that any such extension (which we will also denote by $\mathscr L$) is $f$-ample. (This implies that $\mathscr L \otimes f^*( \mathcal O(n))$ is ample on $\tilde X$ for $n \gg 0$.)
Indeed, let $V = \mathbf P^3 \setminus \{p,q\}$ and let $\tilde V = f^{-1}(V)$. By assumption, $U \cup V = \mathbf P^3$. Moreover, $f|_{\tilde V} \colon \tilde V \to V$ is just the blowup in the disjoint union of curves $C \cup D$. Hence, we have $$\operatorname{Pic}(\tilde V) = f^*\operatorname{Pic}(V) \oplus \mathbf Z [E_C] \oplus \mathbf Z [E_D],$$ where $E_C \to C$ and $E_D \to D$ are the exceptional divisors. A line bundle $f^*\mathscr M \otimes \mathcal O_{\tilde V}(aE_C + bE_D)$ on $\tilde V$ is $f$-ample if and only if $a < 0$ and $b < 0$: clearly these are $f$-ample by construction of the blowup, and conversely the restriction of an $f$-ample to a nontrivial fibre $f^{-1}(x) \cong \mathbf P^1$ for $x \in C \cup D$ should be ample, forcing $a < 0$ and $b < 0$ (since $E_C|_{f^{-1}(x)} = \mathcal O_{\mathbf P^1}(-1)$ for $x \in C$). The same goes for line bundles on $\tilde V \cap \tilde U$. Write $$\mathscr L|_{\tilde V} = f^*\mathscr M \otimes \mathcal O_{\tilde V}(aE_C + bE_D).$$ Restricting this decomposition to $\tilde V \cap \tilde U$ gives $$\mathscr L|_{\tilde V \cap \tilde U} = f^*\left(\mathscr M|_{V \cap U}\right) \otimes \mathcal O_{\tilde V \cap \tilde U}(aE_C + bE_D).$$ Since $\mathscr L|_{\tilde U}$ is $f$-ample, the same goes for $\mathscr L_{\tilde V \cap \tilde U}$, so $a < 0$ and $b < 0$. This in turn implies that $\mathscr L|_{\tilde V}$ is $f$-ample. Since the same holds for $\mathscr L|_{\tilde U}$ by assumption, we conclude that $\mathscr L$ is $f$-ample. $\square$
This gives the required counterexample: let $S = U \subseteq \mathbf P^3$ be an affine open containing $p$ and $q$, and let $S' = U_1 \amalg U_2$ for affine opens $U_1, U_2 \subseteq U$ each containing exactly one of $p$ and $q$ such that $U = U_1 \cup U_2$. (Something like this can be arranged: take $U_1$ the complement of a plane $H_1 \subseteq \mathbf P^3$ through $p$ but not $q$, and $U_2$ the opposite, and let $U$ be the complement of a plane containing $H_1 \cap H_2$ but not $p$ or $q$. Or allow yourself more opens to cover $U$ if you don't want to fidget like this.)
The lemma shows that $f$ is not projective over $U$, and Hironaka showed that $f$ is projective over $U_1$ and $U_2$, hence over $S'$. $\square$
Let $X$ be an algebraic space which is not a scheme, and let $X\to S$ be a smooth proper morphism whose geometric fibres are K3 surfaces. Such data exists with $S $ a finite etale cover of $\mathrm{Spec}\mathbb{Z}[1/n]$ for some large enough $n$.
There is an etale cover $S'\to S$ such that $X_{S'}$ is in fact a scheme. Thus, the answer to your question is no (assuming that you consider morphisms of algebraic spaces $X\to S$).
If you assume $X$ to be a scheme, then probably it is still not true what you want.