It can be two very different things, depending on whether you consider $(\Bbb R^n\setminus \{0\})/\sim$ or $\Bbb R^n/\sim$.
In the former situation, it can indeed be identified with the sphere, as Andrew Maurer pointed out in the comments. One quick way (hit me up, if you want me to elaborate) to see that would be for example by noting that
$$\begin{array}{rcl}
\Bbb S^n & \rightarrow &(\Bbb R^n\setminus\{0\})/\sim\\
x & \mapsto & [x]
\end{array}$$
defines a continuous bijection from a compact space $\Bbb S^n$ to the Hausdorff space $(\Bbb R^n\setminus\{0\})/\sim$, hence a homeomorphism. Continuity is detected by noting that the map is just the composite
$$\Bbb S^n \rightarrow \Bbb R^n\setminus\{0\} \rightarrow (\Bbb R^n\setminus\{0\})/\sim$$
It is bijective by definition and is Hausdorff given the fact that the equivalence relation is induced by a good group action $\Bbb R_{+} \curvearrowright \Bbb R^n\setminus\{0\}$.
In the case of having $\Bbb R^n/\sim$ things get a bit more complicated. Note that by definition of the quotient topology every open subset contains the equivalence class of 0, making it a dense point. In particular this quotient is not Hausdorff, so it won’t be homeomorphic to any of the usual nice spaces.