For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$  For any   smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$  such that  the normal bundle of any submanifold  representing $\alpha$  does not admit  a $spin^c$-structure; see Theorem 3, page 9 of [this paper][1].


If $\alpha^\dagger\in H^4(M,\bZ)$  denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.   

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits   $spin^c$ structures because it admits an almost complex  structure.

**Edit 1.**  A rather deep divisibility theorem shows that  if  $n\geq 3$ and    $E\to S^{2n}$ is a complex vector  bundle, then $c_n(E)\in H^{2n}(S^{2n},\bZ)$ is divisible by $(n-1)!$. 


  [1]: http://arxiv.org/pdf/math/0011178.pdf