Projective after fpqc base change Let $S$ be a Noetherian affine scheme. Let $S'\to S$ be a flat surjective morphism of affine schemes. Let $X\to S$ be a morphism such that $X_{S'}\to S'$ is projective. Is $X\to S$ projective? It is not difficult to see that this holds when $S'\to S$ is a field extension.
The analogous statement for proper morphisms is true because being proper is fpqc local on the base. Hironaka's examples show that the same argument can not work for projective morphisms.
 A: 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 $S'$. $\square$
A: 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.
