fpqc covers of stacks Artin has a theorem (10.1 in Laumon, Moret-Bailly) that if $X$ is a stack which has separated, quasi-compact, 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?
 A: 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 non-empty 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 non-empty, 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 quasi-compact 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.) 
