Suppose that $X$ is a smooth complete positive-dimensional algebraic space over $\mathbb C$. Then $\mathrm H^2(X, \mathbb Q)$ can not be 0. In fact, every algebraic space contains an open dense subscheme. Let $U \subseteq X$ be an open affine subscheme; by a well known-result, the complement $C$ of $U$ has pure codimension 1; consider $C$ as a divisor, with its reduced scheme structure. I claim that the cohomology class of $C$ can not be 0. Let $Y \to X$ a birational morphism, where $Y$ is a smooth projective variety; this exists, by Chow's lemma for algebraic spaces, and resolution of singularities. The pullback of $C$ is a non-zero effective divisor on $Y$, so its cohomology class is not 0. Since formation of cohomology classes of divisors is compatible with pullbacks, this implies the result.

This says very little about the problem of existence of a complex structure. Complete algebraic spaces, or, if you prefer, Moishezon manifolds, are not so distant from projective varieties; it seems clear to me that if $S^6$ has a complex structure, this will be a very exotic animal, very far away from the world of algebraic geometry.

[Edit]: why does $C$ have codimension 1? This is Corollaire 21.12.7 of EGAIV, when $X$ is a scheme. In general you can reduce to the case of a scheme by considering an étale morphism $Y \to X$, where $Y$ is an affine scheme. This map is affine, because $X$ is separated, so the inverse image of $U$ in $Y$ is affine.