Is the pairing induced by the wedge product and integration nondegenerate on de Rham forms?
The Hodge duality gives the non degeneracy of the pairing.
The Hodge star define a duality on de Rham forms, that is, if $\alpha$ is a non-zero $k$-form then $\star \alpha$ is a non-zero $(n-k)$-form. The defining property of the $\star$ operator is indeed given by $\beta \wedge \star \alpha = \langle \alpha, \beta \rangle \operatorname{vol}$, where $\operatorname{vol}$ is a volume form ($M$ is oriented).
Edit (after sitting on it for a while): While the Hodge star gives us the isomorphism between $\Omega^k_{\mathrm{dR}}(M)$ and $\Omega^{n-k}_{\mathrm{dR}}(M)$ (which is what it is explained above), I don't think it gives us an isomorphism of $\Omega^k_{\mathrm{dR}}(M)$ with the dual of $\Omega^{n-k}_{\mathrm{dR}}(M)$, as $\Omega^{n-k}_{\mathrm{dR}}(M)$ is infinite dimensional. So you were correct in the comment, the fact that $\star \star= (-1)^{k(n-k)}$ only guarantees the isomomorphism mentioned above.
Actually, the answer is almost yes on the level of forms themselves. You've provided an inner product on $\Omega^k(M)$, and once you pass to its Hilbert space completion (the so-called space of $L^2$ forms) it is true, by the Riesz representation theorem, that any continuous linear functional $\Omega^k_{L^2}(M) \to \Bbb R$ is uniquely represented by $\alpha \mapsto \langle \alpha, \beta \rangle$; that is, it's uniquely represented by integration against $*\beta$; that is, it's uniquely represented by integration against an $(n-k)$-form. (An $L^2$ form, to be more careful.) To be more precise yet, the map $\Omega^k_{L^2}(M) \to \left(\Omega^k_{L^2}\right)^*$, given by $\alpha \mapsto \langle \cdot,\alpha\rangle$ is an isometry.
The problem with the example you gave, $f \mapsto f(0)$, is that it is not continuous in the $L^2$-topology, so you definitely should not expect it to be given by integration against any kind of form. But given an $L^2$-continuous functional $\Omega^k(M) \to \Bbb R$, you know it's given by integration against an $(n-k)$-form - but only necessarily an $L^2$ one, as an artifact of the non-completeness of $\Omega^k(M)$.