If a subset $X\subset R^3$ is closed, locally path connected and semilocally simply connected, then for each $x\in X$, the group $\pi_1(X,x)$ is torsion-free. See my answer here. I have no idea what happens if any of these conditions is weakened.

Proposition. Suppose that a subset $X\subset R^3$ is semilocally simply connected (SLSC). Then for each $x\in X$, the group $G= \pi_1(X,x)$ is torsion-free.

Proof. I will be using the fact that if $U$ is an open connected subset of $R^3$, then $\pi_1(U)$ is torsion-free, see for instance here.

Let $c: S^1\to X$ be a loop representing an element of order $n$ in $\pi_1(X,x)$. Accordingly, let $c_n: S^1\to X$ be the precomposition of $c$ with the map $z\mapsto z^n$ and $h: D^2\to X$ be an extension of $c_n$ to the unit disk. The image $Y:= h(D^2)$ is compact and locally path-connected. Thus, there is a function $\phi(\delta)$ such that if $d(y_1, y_2)\le \delta$, $y_1, y_2\in Y$, then there is a path of diameter $\le \phi(\delta)$ in $Y$ connecting $y_1, y_2$. Furthermore, since $Y$ is compact and $X$ is assumed to be SLSC, there exists $\epsilon>0$ such that if $\alpha: S^1\to Y$ has image of diameter $\le \epsilon$, then $\alpha$ extends to a continuous map $D^2\to X$.

Now, consider the system of open $\frac{1}{i}$-neighborhoods $U_n$ of $Y$ in $R^3$. Let $r: U_i\to Y$ denote a (likely discontinuous) nearest-point projection (defined via the Axiom of Choice). Since each group $\pi_1(U_i,x)$ is torsion-free, the map $c: S^1\to Y$ extends to a (continuous) map $f_i: D^2\to U_i$.

I will now imitate the standard argument (I think, due to Borsuk), from the proof that each finite-dimensional compact locally-contractible metrizable space is ANR.

Given a triangulation $T$ of $D^2$, I define the map $g_i: T^{(0)}\to Y$ as the composition of the restriction of $f_i$ to the $T^{(0)}$ with the projection $r$. The goal is to show that for large $i$, the map $g_i$ extends to a map $D^2\to X$ which restricts to $c$ on $S^1=\partial D^2$.

First of all, if $\frac{1}{i}\le \delta$, and $T$ is such that the diameters of the images under $f_i$ of the edges of $T$ are $\le\delta$, then for each edge $e=[v,w]$ of $T$, $d(g_i(v), g_i(w))\le 3\delta$. Hence, by the local path connectivity of $Y$, we can extend $g_i$ to $e$ so that $g_i(e)\subset Y$ has diameter $\le \phi(3\delta)$. In the case of boundary edges of the disk $D^2$, we will assume that $g_i|_e=c|_e$. Note that for each 2-simplex $\Delta$ in $T$, the diameter of $g_i(\partial \Delta)$ is $\le 3\phi(3\delta)$. By taking $i$ sufficiently large and taking the triangulation $T$ sufficiently fine, we can assume that $3\phi(3\delta)\le \epsilon$, where $\epsilon$ is defined as above. Hence, for this value of $i$, the map $g_i$ extends to a map $g: D^2\to X$. It follows that $[c]=1\in G=\pi_1(X,x)$ and, hence, $G$ is torsion-free. qed