Let $n >1$ and $\omega \in \Omega^{n-1}(\mathbb{R}^{n+1}\setminus\{0\})$ such that $d\omega = 0$. Is the following statement true:
For any compact, oriented, $(n-1)$-dimensional submanifold $M$ of $\mathbb{R}^{n+1}\setminus\{0\}$, we have $\displaystyle \int\limits_{M}j^*\omega = 0$ with $j:M \rightarrow \mathbb{R}^{n+1}\setminus\{0\}$ the inclusion map.
This calls for the Theorem of Stokes ( I thought to apply it two times), but $M$ isn't necessarily a boundary of another Manifold.
On
Consider n=2 and let $X=(x,y,z)/r^3$ where $r=\sqrt{x^2+y^2+z^2}$, $X$ is a vector field defined on $\mathbb{R}^3-\{0\}$. It's easy to compute $$\operatorname{div}(X)=\sum\frac{1\cdot r^3-x\cdot3r^2\cdot r_x}{r^6}$$ $$=\sum\frac{r^3-x\cdot 3r^2 (x/r)}{r^6}$$ $$=\sum\frac{r^3-3rx^2}{r^6}$$ $$=\frac{3r^3}{r^6}-\frac{3r}{r^6}\sum x^2$$ $$=3/r^3-3/r^3=0$$ So let $\Omega=dx\wedge dy\wedge dz$ and $\omega=i(X)\Omega$ we have $d\omega=\operatorname{div}(X)\Omega=0$ but $\int_{S^2}j^*\omega=4\pi$
$\mathbb{R}^{n+1} \setminus \lbrace 0 \rbrace$ is homotopy equivalent to $\mathbb{S}^n$, and so in particular the $(n-1)$-st cohomology group $$ H^{n-1}(\mathbb{R}^{n+1} \setminus \lbrace 0 \rbrace) = H^{n-1}(\mathbb{S}^n) $$ vanishes. In particular, your closed form $\omega$ is exact; that is, there is an $(n-2)$-form $\eta$ with $d\eta = \omega$. You can now use Stokes' theorem in the usual way, together with the fact that $\partial M = \emptyset$, to show that your integral is $0$.