Is the next statement true?
Let $M$ be a non-compact linearly connected oriented topological manifold of dimension $n$, and let $M^+$ be the one-point compactification of $M$. Then there is a canonical isomorphism $$ H^q(M;\mathbb{Z})\cong\widetilde{H}_{n-q}(M^+;\mathbb{Z}) $$ where $0\leqslant q\leqslant n$ and $\widetilde{H}$ means the reduced homology.
If this is true, where could I find a proof?
As pointed out in the comments, this sometimes fails if $M$ is not the homotopy type of a compact manifold with boundary, but is true if $M \cong \mathrm{int}(\bar{M})$ for a compact, orientable $n$-manifold $\bar{M}$. Here is a short argument which derives both classical Poincare duality and this version at the same time.
Recall that for a finite space $X$, the Spanier-Whitehead dual $X_+^\vee$ is given by the function spectrum $F(\Sigma^\infty_+ X,S^0)$. This spectrum has the property that $H_*(X)\cong H^{-*}(X_+^\vee)$ and $H^*(X) \cong H_{-*}(X_+ ^\vee)$. Note the right hand sides are "reduced" because they are the (co)homology of a spectrum
Notice that $M^+ \cong \bar{M}/\partial \bar{M}$, so it suffices to show $H^q(\bar{M}) \cong \bar{H}_{n-q}(\bar{M}/\partial \bar{M})$. Let $i:\bar{M} \rightarrow \mathbb{R}^N$ be a codimension $c$ embedding with a disk bundle neighborhood $\nu$. Dolde-Puppe proved that in these circumstances $\Sigma^{n+c}M_+^\vee \simeq \Sigma^\infty \bar{M}^{\nu}/\partial\bar{M}^{\nu|_{\partial M}}$, where $\bar{M}^\nu$ denotes the Thom complex of $\nu$. By (relative) Thom isomorphism, the latter has the (co)homology of $\Sigma^c (\bar{M}/\partial \bar{M})$.
One can now apply (co)homology to this equivalence, and then apply the (co)homological Thom isomorphism. Then we get two statements of Poincare duality:
$H^{n+c-*}(\bar{M})\cong \bar{H}_{*-c}(\bar{M}/\partial\bar{M})$
$H_{n+c-*}(\bar{M})\cong \bar{H}^{*-c}(\bar{M}/\partial\bar{M})$
Letting $n+c-*=q$, the first reduces to $H^q(\bar{M})\cong \bar{H}_{n-q}(\bar{M}/\partial \bar{M}))$ and the second reduces to classical Poincare duality, as desired. A variation of this argument should work in the slightly more general case where $M$ is homotopy equivalent to a finite complex, but not necessarily the interior of a closed manifold.
