You can certainly show that many such space forms do not embed. The simplest version would be that a lens space ($L =S^n/\mathbb{Z}_k$; here $n$ should be odd) whose fundamental group has order $k=p^r$ for $p$ a prime does not embed. To see this, replace $\mathbb{R}^{n+1}$ with $S^{n+1}$ that would be split into two pieces, say $A$ and $B$, by $L$. Then the homology of $L$ is the direct sum of the homology groups of $A$ and $B$ by an easy Mayer-Vietoris argument. In particular, either $A$ or $B$ would be a null-bordism over $\mathbb{Z}_k$ for $L$, which represents the generator of $H_n(B\mathbb{Z}_k) = \mathbb{Z}_k$. The observation about the homology goes back quite a while; cf. Hantzsche, W.; Einlagerung von Mannigfaltigkeiten in euklidische Räume. Math. Z. 43 (1938), no. 1, 38–58.

I imagine that this would work for composite $k$ with a slightly more elaborate argument. There are also lots of other obstructions; unlike in the classical case $S^n/\Gamma$ might not be stably parallelizable, so there is no codimension-one embedding. In fact, it's hard to find stably parallelizable lens spaces; cf. Ewing, John; Moolgavkar, Suresh; Smith, Larry; Stong, R. E. Stable parallelizability of lens spaces. J. Pure Appl. Algebra 10 (1977/78), no. 2, 177–191.