Realizing cohomology classes by submanifolds

Your question is just a reformulation of what Thom did, so the answer is always yes.

Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map $$ \Omega_\bullet(M)\otimes \Bbb R \to H^{\bullet}_{DR}(M;\Bbb R)^\ast $$ is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^\bullet_{DR}(M;\Bbb R)^\ast \cong H_\bullet(M;\Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which $$ \Omega_\bullet(M) \to H_\bullet(M) $$ is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $\Sigma\to M$ the image of the fundamental class $[\Sigma]$ in $H_\bullet(M)$.

It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map $$\pi_\bullet^{\text{st}}(M)\to \Omega_\bullet(M)$$ from stable homotopy to oriented bordism. The composite $$\pi_\bullet^{\text{st}}(M)\to \Omega_\bullet(M)\to H_\bullet(M) $$ is just the usual Hurewicz map.

It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $S\to h\Bbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $\Omega_\bullet(M)\to H_\bullet(M)$ is a surjection modulo torsion.