In complex dimension 3 or more it is still an open conjecture
(which was re-stated Yau a couple of years ago in his UCLA lectures).
There is not a single known example of an almost complex manifold
of dimension $\geq$ 3 not admitting a complex structure.
In dimension 2 it is easy, of course, because
the non-Kähler complex surfaces are understood
much better than Kähler ones: every non-Kähler
surface with $b_1 >1$ is diffeomorphic to a blow-up of a
locally trivial elliptic fibration over a curve. Hence any
4-dimensional compact almost complex manifold
with odd $b_1 >1$ and a fundamental group not virtually
isomorphic (*) to a cross-product of a fundamental group of a curve
and $\mathbb{Z}$, cannot be a complex surface.
(*) Here "virtually isomorphic" means "isomorphic up to a
finite index subgroup".
Assume that $M$ is oriented throughout. Recall that $M$ has a $\text{Spin}^c$ structure iff the third integral Stiefel-Whitney class $\beta w_2 = W_3 \in H^3(M, \mathbb{Z})$ is trivial. Actually more is true: $\text{Spin}^c$ structures on $M$ are in bijection with trivializations of $W_3$, which are a torsor over $H^2(M, \mathbb{Z})$. So we get a functorial way to associate a $\text{Spin}^c$ structure to an almost complex structure for every choice of nullhomotopy of the composite map
$$BU(n) \to BSO(2n) \xrightarrow{W_3} B^3 \mathbb{Z}.$$
Isomorphism classes of nullhomotopies are a torsor over $H^2(BU(n), \mathbb{Z}) \cong \mathbb{Z}$. I don't know in what sense there's a canonical one, but once you pick one you don't have to make any further choices, and in particular you don't have to make any further choices involving $M$. (Edit #2: But see below.)
As for almost complex structures, there is a canonical fiber bundle
$$SO(2n)/U(n) \to BU(n) \to BSO(2n)$$
and almost complex structures on $M$ correspond to sections of the pullback of this bundle along the classifying map $M \to BSO(2n)$ of the tangent bundle. This admits a map of bundles to the corresponding bundle
$$B^2 \mathbb{Z} \to B \text{Spin}^c(2n) \to BSO(2n)$$
describing $\text{Spin}^c$ structures. The map $SO(2n)/U(n) \to B^2 \mathbb{Z}$ describes a canonical line bundle on the space of linear complex structures on $\mathbb{R}^{2n}$, namely the complex determinant bundle.
This tells us that we already can't expect all $\text{Spin}^c$ structures to come from almost complex structures when $n = 1$: here $SO(2)/U(1)$ is contractible, so there is in a very strong sense a unique almost complex structure on an oriented surface, but $\text{Spin}^c$ structures are a torsor over $H^2(-, \mathbb{Z})$, which can be nontrivial, e.g. on an oriented closed surface.
Edit #2: I believe that there is in fact a distinguished nullhomotopy of the composite map $BU(n) \xrightarrow{W_3} B^3 \mathbb{Z}$, as follows.
We now need the additional assumption that $M$ is Riemannian (although the choice of Riemannian metric won't matter). Then a conceptual description of $W_3 \in H^3(M, \mathbb{Z})$ is that it classifies the bundle of complex Clifford algebras $\text{Cliff}(T_x(M)) \otimes \mathbb{C}$ up to Morita equivalence. Trivializations of this bundle up to Morita equivalence correspond to Clifford module bundles with fiber the unique irreducible representation of $\text{Cliff}(2n) \otimes \mathbb{C}$ (complex spinor bundles).
If $M$ has an almost complex structure, then a distinguished complex spinor bundle can be constructed out of the complex exterior algebra of the tangent bundle. See Exercise 2.1.37 in Freed's Geometry of Dirac Operators for details. The torsor structure over $H^2(M, \mathbb{Z})$ comes from tensoring this bundle with complex line bundles on $M$, and the action of $H^2(BU(n), \mathbb{Z})$ comes from tensoring this bundle with powers of the canonical bundle. It's possible one might want to do this with a particular power; I'm not sure what's going on here exactly, but see Exercise 2.1.54 in Freed.
Edit #3: Now that the third question has been clarified, I don't know the answer off the top of my head. In the case of closed oriented surfaces we know that there's a $\mathbb{Z}$'s worth of $\text{Spin}^c$ structures. The unique almost complex structure gives rise to one, and the $2^{2g}$ spin structures also give rise to one (not depending on the spin structure, basically because the Bockstein homomorphism $H^1(M, \mathbb{Z}_2) \to H^2(M, \mathbb{Z})$ is zero in this case). But I don't know if they're the same one; they might differ by a power of the canonical bundle or something (or not, depending on your convention for trivializing $W_3$ as above). In summary,
- Yes, more or less,
- No, and
- I don't know, but I would guess no.
Best Answer
This is still an open problem. See this paper for some progress, which was prompted by this MO question.