It's not exact. To see this, as suggested by Andrea, we use the spherical coordinates:
$$(x_1,x_2,x_3)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),0<\theta<2\pi,0<\phi<\pi.$$
Restricted to $\mathbb{S}^2$, we have $r=1$, i.e.
$$(x_1,x_2,x_3)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$
which implies that
$$dx_1|_{\mathbb{S}^2}=-\sin\phi\sin\theta\,d\theta+\cos\phi\cos\theta\,d\phi,$$
$$dx_2|_{\mathbb{S}^2}=\sin\phi\cos\theta\,d\theta+\cos\phi\sin\theta\,d\phi,$$
$$dx_3|_{\mathbb{S}^2}=-\sin\phi\,d\phi.$$
This gives
$$dx_1\wedge dx_2\big|_{\mathbb{S}^2}=-\sin\phi\cos\phi\,d\theta\wedge d\phi,$$
$$dx_2\wedge dx_3\big|_{\mathbb{S}^2}=-\sin^2\phi\cos\theta\,d\theta\wedge d\phi,$$
$$dx_3\wedge dx_1\big|_{\mathbb{S}^2}=-\sin^2\phi\sin\theta\,d\theta\wedge d\phi.$$
Combining all these, we can express $\sigma$ in terms of spherical coordinates:
$$\sigma\big|_{\mathbb{S}^2}=\Big(\frac{x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2}{(x_1^2+x_2^2+x_3^2)^{3/2}}\Big)\Big|_{\mathbb{S}^2}$$
$$=\big(x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2\big)\big|_{\mathbb{S}^2}$$
$$=-\sin^3\phi\cos^2\theta\,d\theta\wedge d\phi-\sin^3\phi\sin^2\theta\,d\theta\wedge d\phi-\sin\phi\cos^2\phi\,d\theta\wedge d\phi$$
$$=-\sin\phi\,d\theta\wedge d\phi.$$
Suppose that $\sigma$ is exact, then $\sigma=df$ for some one-form $f$. By Stoke's theorem, we have
$$\int_{\mathbb{S}^2}\sigma=\int_{\mathbb{S}^2}df=\int_{\partial \mathbb{S}^2}f=0$$
where the last equality follows from the fact that $\partial \mathbb{S}^2=\varnothing$, i.e. $\mathbb{S}^2$ is an manifold with no boundary. However, from the above calculation,
$$\int_{\mathbb{S}^2}\sigma=-\int_{0}^\pi\int_0^{2\pi}\sin\phi\,d\theta d\phi=-4\pi, $$
which is a contradiction.
