Artin has a theorem (10.1 in Laumon, MoretBailly) that if $X$ is a stack which has separated, quasicompact, representable diagonal and an fppf cover by a scheme, then $X$ is algebraic. Is there a version of this theorem that holds for fpqc covers?

It is false. I'm not sure what the comment about algebraic spaces has to do with the question, since algebraic spaces do admit an fpqc (even \'etale) cover by a scheme. This is analogous to the fact that the failure of smoothness for automorphism schemes of geometric points is not an obstruction to being an Artin stack. For example, $B\mu_n$ is an Artin stack over $\mathbf{Z}$ even though $\mu_n$ is not smooth over $\mathbf{Z}$ when $n > 1$. Undeterred by this, we'll make a counterexample using $BG$ for an affine group scheme that is fpqc but not fppf. First, we set up the framework for the counterexample in some generality before we make a specific counterexample. Let $S$ be a scheme and $G \rightarrow S$ a $S$group whose structural morphism is affine. (For example, if $S = {\rm{Spec}}(k)$ for a field $k$ then $G$ is just an affine $k$group scheme.) If $X$ is any $G$torsor for the fpqc topology over an $S$scheme $T$ then the structural map $X \rightarrow T$ is affine (since it becomes so over a cover of $T$ that splits the torsor). Hence, the fibered category $BG$ of $G$torsors for the fpqc topology (on the category of schemes over $S$) satisfies effective descent for the fpqc topology, due to the affineness requirement. The diagonal $BG \rightarrow BG \times_S BG$ is represented by affine morphisms since for any pair of $G$torsors $X$ and $Y$ (for the fpqc topology) over an $S$scheme $T$, the functor ${\rm{Isom}}(X,Y)$ on $T$schemes is represented by a scheme affine over $T$. Indeed, this functor is visibly an fpqc sheaf, so to check the claim we can work locally and thereby reduced to the case $X = Y = G_T$ which is clear. Now impose the assumption (not yet used above) that $G \rightarrow S$ is fpqc. In this case I claim that the map $S \rightarrow BG$ corresponding to the trivial torsor is an fpqc cover. For any $S$scheme $T$ and $G$torsor $X$ over $T$ for the fpqc topology, the functor $$S \times_{BG} T = {\rm{Isom}}(G_T,X)$$ on $T$schemes is not only represented by a scheme affine over $T$ (namely, $X$) but actually one that is an fpqc cover of $T$. Indeed, to check this we can work locally over $T$, so passing to a cover that splits the torsor reduces us to the case of the trivial $G$torsor over the base (still denoted $T$), for which the representing object is $G_T$. So far so good: such examples satisfy all of the hypotheses, and we just have to prove in some example that it violates the conclusion, which is to say that it does not admit a smooth cover by a scheme. Take $S = {\rm{Spec}}(k)$ for a field $k$, and let $k_s/k$ be a separable closure and $\Gamma = {\rm{Gal}}(k_s/k)^{\rm{opp}}$. (The "opposite" is due to my implicit convention to use left torsors on the geometric side.) Let $G$ be the affine $k$group that "corresponds" to the profinite group $\Gamma$ (i.e., it is the inverse limit of the finite constant $k$groups $\Gamma/N$ for open normal $N$ in $\Gamma$). To get a handle on $G$torsors, the key point is to give a more concrete description of the ``points'' of $BG$. Claim: If $A$ is a $k$algebra and $B$ is an $A$algebra, then to give a $G$torsor structure to ${\rm{Spec}}(B)$ over ${\rm{Spec}}(A)$ is the same as to give a right $\Gamma$action on the $A$algebra $B$ that is continuous for the discrete topology such that for each open normal subgroup $N \subseteq \Gamma$ the $A$subalgebra $B^N$ is a right $\Gamma/N$torsor (for the fpqc topology, and then equivalently the \'etale topology). Proof: Descent theory and a calculation for the trivial torsor. QED Claim Example: $A = k$, $B = k_s$, and the usual (right) action by $\Gamma$. Corollary: If $A$ is a strictly henselian local ring then every $G$torsor over $A$ for the fpqc topology is trivial. Proof: Let ${\rm{Spec}}(B)$ be such a torsor. By the Claim, for each open normal subgroup $N$ in $\Gamma$, $B^N$ is a $\Gamma/N$torsor over $A$. Since $A$ is strictly henselian, this latter torsor is trivial for each $N$. That is, there is a $\Gamma/N$invariant section $B^N \rightarrow A$. The nonempty set of these is finite for each $N$, so by set theory nonsense with inverse limits of finite sets (ultimately not so fancy if we take $k$ for which there are only countably many open subgroups of $\Gamma$) we get a $\Gamma$invariant section $B \rightarrow A$. QED Corollary Now suppose there is a smooth cover $Y \rightarrow BG$ by a scheme. In particular, $Y$ is nonempty, so we may choose an open affine $U$ in $Y$. I claim that $U \rightarrow BG$ is also surjective. To see this, pick any $y \in U$ and consider the resulting composite map $${\rm{Spec}} \mathcal{O}_{Y,y}^{\rm{sh}} \rightarrow BG$$ over $k$. By the Corollary, this corresponds to a trivial $G$torsor, so it factors through the canonical map ${\rm{Spec}}(k) \rightarrow BG$ corresponding to the trivial $G$torsor. This latter map is surjective, so the assertion follows. Hence, we may replace $Y$ with $U$ to arrange that $Y$ is affine. (All we just showed is that $BG$ is a quasicompact Artin stack, if it is an Artin stack at all, hardly a surprise in view of the (fpqc!) cover by ${\rm{Spec}}(k)$.) OK, so with a smooth cover $Y \rightarrow BG$ by an affine scheme, the fiber product $$Y' = {\rm{Spec}}(k) \times_{BG} Y$$ (using the canonical covering map for the first factor) is an affine scheme since we saw that $BG$ has affine diagonal. Let $A$ and $B$ be the respective coordinate rings of $Y$ and $Y'$, so by the Claim there is a natural $\Gamma$action on $B$ over $A$ such that the $A$subalgebras $B^N$ for open normal subgroups $N \subseteq \Gamma$ exhaust $B$ and each $B^N$ is a $\Gamma/N$torsor over $A$. But $Y' \rightarrow {\rm{Spec}}(k)$ is smooth, and in particular locally of finite type, so $B$ is finitely generated as a $k$algebra. Since the $B^N$'s are $k$subalgebras of $B$ which exhaust it, we conclude that $B = B^N$ for sufficiently small $N$. This forces such $N$ to equal $\Gamma$, which is to say that $\Gamma$ is finite. Thus, any $k$ with infinite Galois group does the job. (In other words, if $k$ is neither separably closed nor real closed, then $BG$ is a counterexample.) 

