No, the Grigorchuk group, which is torsion, admits a 2-ended connected Schreier graph. See for instance http://www.math.tamu.edu/~yvorobet/Research/Schreier.pdf

*Edit:* Here's a more elementary example, just assuming the bare existence of an infinite finitely generated torsion group, and also produces examples with bounded torsion, and examples with an infinite space of ends.

Let $\Gamma,C$ be any finitely generated groups (in a first reading, one can assume that $\Gamma$ is 1-ended and that $|C|=2$). We consider the Schreier graph $X$ of the wreath product $G=C\wr\Gamma=\Gamma\ltimes C^{(\Gamma)}$ with respect to the subgroup $C^{(\Gamma\smallsetminus\{1\})}$. We can view $X$ as the product $\Gamma\times C$, where $\Gamma$ and $C$ act as
$$g\cdot (g',c')=(gg',c');\quad c\cdot (g',c')=(g',c') \;(g\neq 1)\quad c\cdot (1,c')=(1,cc').$$

One fixes finite generating subsets $S_\Gamma,S_C$, so that $S=S_\Gamma\cup S_C$ generates the wreath product $G$.

If $|C|=2$, the Schreier graph $X$ can therefore be described as the disjoint union of two copies of the Cayley graph of $G$, joined by a single edge joining their basepoints. If $G$ is 1-ended, this is 2-ended.

In general ($C$ arbitrary) $X$ is obtained from the Cayley graph of $C$ by "planting" Cayley graphs of $G$ at each point.

In particular, say when $C$ has finitely many ends, the number of ends of $X$ is equal to $|C|$ times the number of ends of $G$. This can be chosen to be infinite, say when both $C$ and $\Gamma$ are both infinite Burnside groups, in which case $G$ even has bounded torsion.

If $\Gamma,C$ are both finitely-ended, the space of ends of $X$ is countable anyway: if $C$ is infinite with 1 or 2 ends, then this is just a 1-point or 2-point compactification of an infinite countable set.

So, when $G$ is torsion, the resulting space of ends we thus obtain are all finite set, and the 1-point compactification of an infinite countable set.

In general, this can also produce $X$ to have a space of ends that is uncountable with isolated points (e.g., when $\Gamma=\mathbf{Z}$ and $C$ is $\infty$-ended).