Being "affine" in this case does not make much sense,
because the hyperkaehler deformation is a complex manifold, without
a fixed algebraic structure. Simpson produced an example of a
hyperkaehler deformation of a space of flat bundles
admitting several algebraic structures, both
inducing the same Stein complex structure; one of them
is affine, another has no global algebraic functions.
In fact, the space F of flat line bundles on elliptic curve
(with an appropriate algebraic structure, defined by
Simpson) is an example of such a manifold,
it is biholomorphic to $C^*\times C^*$, but
this biholomorphic equivalence is not algebraic,
and F has no global algebraic functions.
However, you can show that a hyperkaehler deformation
of a resolution of something affine has no non-trivial complex
subvarieties (arXiv:math/0312520), except, possibly, some
hyperkaehler subvarieties The latter don't exist, because
the holomorphic symplectic form $\Omega$ on such a manifold
is is lifted from the base, which is affine, hence $\Omega$
vanishes on all complex subvarieties.
Therefore, a typical fiber of such a deformation is Stein.
Indeed, a hyperkaehler deformation of a
resolution of something affine remains holomorphically convex. To see this
if you produce a function which is strictly plurisubharmonic outside of
a compact set (we have such a function, because we started from something
affine), and apply the Remmert reduction.
So here's what I think should work (it is based on my comments above). I'm going to assume for simplicity that the variety means irreducible (just to avoid some silly complications, it is not really needed).
Start with $f : Z \to X$ projective and birational (as discussed above, $Z$ is contructed via Chow's Lemma). We don't know that $X$ is quasi-projective, so we can't apply Hartshorne II, 7.11 and thus immediately argue that $Z$ is itself a blow-up.
Set $U_i$ to be an open affine (or even quasi-projective) cover of $X$ and fix $V_i = f^{-1}(U_i)$. On each chart $U_i$, set $J_i$ to be an ideal sheaf such that $f_i : V_i \to U_i$ is the blow-up of $J_i$.
For each $i$, fix $I_i$ to be a (generically non-zero) ideal sheaf on $X$ such that $I_i |_{U_i} = J_i$.
Now fix $I = \prod_i I_i$. This is an ideal sheaf on $X$. On each open chart $U_i$, it is equal to $J_i \cdot \prod_{j \neq i} I_j|_{U_i}$.
Set $\pi : Y \to X$ to be the blow-up of $I$. We wish to show that $\pi$ factors through $f$, and is in fact a blow-up of $Z$, and so $Y$ is indeed projective. Set $W_i = \pi^{-1}(U_i)$.
All schemes involved are separated, and so to verify that $Y \to X$ actually factors through $Z \to X$, it is sufficient to work locally, so work on a chart $U_i$.
There we are comparing the blow-up of $J_i$ with the blow-up of $J_i \cdot \prod_{j \neq i} I_j|_{U_i} = J_i \cdot K_i$. $V_i$ is the blow-up of $J_i$ and $W_i$ is the blow-up of $J_i \cdot K_i$. However, I claim it is a straightforward exercise to verify that $W_i$ is the same as the blow-up of the ideal sheaf $(J_i \cdot K_i) \cdot \mathcal{O}_{V_i}$.
Let me give a hint as to how to do this.
Set $R = \Gamma(U_i, \mathcal{O}_X)$, set $J_i = (x_1, \dots, x_n)$ and $K_i = (y_1, \dots, y_m)$. Then the blow-up of $J_i$ is covered by affine charts $U_{i,l} = \text{Spec} R[x_1/x_l, \dots, x_n/x_l]$ where the gluings are the obvious ones. Likewise the blowup of $J_i \cdot K_i$ is covered by charts $U_{i,s,t} = \text{Spec} R[(x_1y_1)/(x_s y_t), \dots, (x_1y_m)/(x_s y_t), \dots, (x_ny_m)/(x_s y_t) ]$. Which are easily checked to be the blow-ups of $U_{i,s}$ at $(J_i \cdot K_i) \cdot R[x_1/x_l, \dots, x_n/x_l]$ (unless I've completely forgotten how to do this).
This sort of computation should be viewed as a generalization of the fact that the blow-up of an ideal is the same as the blow-up of a power of an ideal.
But this proves everything I claimed, right? $Y$ is exactly the blow-up of $Z$ at $I \cdot \mathcal{O}_Z$, and so $Y$ is projective, and is the blow-up of some ideal sheaf.
Of course, writing this as a sequence of blow-ups at subvarieties (and not subschemes/ideals) is probably much much harder.
Best Answer
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.