First, there is no comprehensive list of all models of modular curves of genus $1$. (There is a list of congruence subgroups of $SL_{2}(\mathbb{Z})$ here.) Many cases have been computed, including the $X_{0}(n)$, $X_{1}(n)$, $X(n)$, as well as many cases that have prime power level. (With David Zureick-Brown, I did all of the cases whose level is a power of $2$ that might have rational points. See our paper here). Another really handy tool is the code that David Zywina has written (see the paper here) that will count points on any modular curve $X_{G}$ (provided $G \subseteq GL_{2}(\mathbb{Z}/n\mathbb{Z})$ has surjective determinant). This will allow one to determine the isogeny class of ${\rm Jac}~X_{G}$, but this is not quite enough to determine the $j$-invariant.

Based on the examples that are known, it is not clear to me there is a relationship between the integrality of the $j$-invariant and the torsion structure. Here are some examples:

$X_{0}(11): y^{2} + y = x^{3} - x^{2} - 10x - 20$, $j$-invariant is $-122023936/161051$ is not integral, but the subgroup $\Gamma_{0}(11)$ is torsion-free (well, its image in ${\rm PSL}_{2}(\mathbb{Z})$ is).

$X_{0}(17): y^{2} + xy + y = x^{3} - x^{2} - x - 14$, $j$-invariant is $-35937/83521$ is not integral, and the subgroup $\Gamma_{0}(17)$ is not torsion-free (because it has elliptic elements of order $2$).

$X_{0}(27): y^{2} + y = x^{3} - 7$, $j$-invariant is zero (and the curve is isomorphic to the Fermat cubic). The subgroup is torsion-free.

$X_{0}(49): y^{2} + xy = x^{3} - x^{2} - 2x - 1$, $j$-invariant is $-3375$ (and has CM by the ring of integers in $\mathbb{Q}(\sqrt{-7})$). The subgroup is not torsion-free, as it contains elliptic elements of order $3$.

Many of these curves have CM. ($X_{0}(32)$, $X_{0}(36)$, $X_{0}(49)$, $X_{{\rm ns}}^{+}(11)$, $X(6)$, most, but not all of the $2$-power level genus $1$ curves.)