I don't know who first asked this question, but it's a question that I think many differential and complex geometers have tried to answer because it sounds so simple and fundamental. There are even a number of published proofs that are not taken seriously, even though nobody seems to know exactly why they are wrong.

Of course, I'm not about to answer this question one way or the other, but there are at least a couple of interesting things one might point out. Firstly, it has been shown (although I forget by whom) that there is no complex structure on S^{6} which is also orthogonal with respect to the round metric. The proof uses twistor theory. The twistor space of S^{6} is the bundle whose fibre at a point p is the space of orthogonal almost complex structures on the tangent space at p. It turns out that the total space is a smooth quadric hypersurface Q in CP^{7}. If I remember rightly, an orthogonal complex structure would correspond to a section of this bundle which is also complex submanifold of Q. Studying the complex geometry of Q allows you to show this can't happen. Secondly, there is a related question: does there exist a nonstandard complex structure on CP^{3}? To see the link, suppose there is a complex structure on S^{6} and blow up a point. This gives a complex manifold diffeomorphic to CP^{3}, but with a nonstandard complex structure, which would seem quite a weird phenomenon. On the other hand, so little is known about complex threefolds (in particular those which are not Kahler) that it's hard to decide what's weird and what isn't. Finally, I once heard a talk by Yau which suggested the following ambitious strategy for finding complex structures on 6manifolds. Assume we are working with a 6manifold which has an almost complex structure (e.g. S^{6}). Since the tangent bundle is a complex vector bundle it is pulled back from some complex Grassmanian via a classifying map. Requiring the structure to be integrable corresponds to a certain PDE for this map. One could then attempt to deform the map (via a cunning flow, continuity method etc.) to try and solve the PDE. I have no idea if anyone has actually tried to carry out part of this program. 


A little more detail to Joel's first paragraph (I can't see how to add a comment to it, sorry!). The argument that there is no orthogonal complex structure on the 6sphere is due to Claude Lebrun and the point is that such a thing, viewed as a section of twistor space, has as image a complex submanifold. Now, on the one hand, this submanifold is Kaehler, and so has nontrivial second cohomology, since the twistor space is Kaehler. On the other hand, the section itself provides a diffeomorphism of our submanifold with the 6sphere which has trivial second cohomology. Neat, huh? 


If such a complex structure exists, it would weird indeed! For example, as shown by Campana, Demailly and Peternell (Compositio 112, 7791), if such a thing exists, then $S^6$ would have no nonconstant meromorphic functions. In particular, $S^6$ can't be Moishezon, let alone algebraic. 


Continuing Joel Fine and Fran Burstall's answer about, indeed "neat", Lebrun's result. Just want to recall that the "orthogonal" twistor space of any $2n$dimensional pseudosphere $SO(2p+1,2q)/SO(2p,2q)$ can be written as $SO(2p+2,2q)/U(p+1,q)$. So the Kähler manifold in question, in case of the 6sphere, is $SO(8)/U(4)$. One should think of each $j:T_xS^6\rightarrow T_xS^6$ as a linear map on $R^8$ with $j(x)=1$ and $j(1)=x$. Well, proofs have been rewritten of LeBrun's result. I wish I had more opinion on this: http://arxiv.org/abs/math/0509442 


Here is a shot in the dark (Disclosure: I really know nothing about this problem). Let $G:=\mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$g\cdot(a,b,c)=(gag^{1},gbg^{1},gcg^{1}).$$ Then the quotient space is homeomorphic to $S^6$ (see BratholdtCooper). The evaluation map shows that the character variety $\mathfrak{X}:=\mathrm{Hom}(\pi_1(\Sigma),G)/G$ is homeomorphic to $G^3/G,$ where $\Sigma$ is an elliptic curve with two punctures. Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over $\Sigma$ (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties. Letting the boundary data vary over all possibilities gives a foliation of $\mathfrak{X}\cong G^3/G\cong S^6$. Therefore, we have a foliation of $S^6$ where generic leaves are projective varieties; in particular, complex. Moreover, the leaves are symplectic given by Goldman's 2form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of $\mathfrak{X}$.
Here are some issues:
Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question. 


This is a famous openproblem. It is still unknown. 


Personally, I do not think that that proof is correct. This is a simple question of a compact homogeneous spaces. Any even dimensional compact Lie group is a (homogeneous) complex torus bundle over a projective rational homogeneous space (which is also simply connectedK\"ahlerEinstein with positive Ricci curvature) and therefore is complex. The paper basically said that the complex structure J_H comes down to S^6 is integrable. His reason was that J_H is the restriction of J_{G_2} to H. However, H is not closed under the Lie bracket. That is why J_H can not simply come down to S^6. 

