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, $Bmu_n$ 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.)
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