Let $F_n$ be free on $n$ letters and $g_1,...,g_{2m}\in F_n$. Can $F_n/⟨⟨[g_1,g_2],...,[g_{2m-1},g_{2m}]⟩⟩$ have torsion elements? 
Let $F_n$ be the free group on $n$ letters.  Let $g_1,...,g_{2m} \in F_n$, can the group $$F_n / \langle\langle[g_1,g_2],...,[g_{2m-1},g_{2m}]\rangle\rangle$$ ever have torsion elements?  

The double angle brackets means "normal subgroup generated by" and $[a,b] = aba^{-1}b^{-1}$.  

This problem arose when I saw a question in an old paper of Yanagawa ("On Ribbon 2-knots, II") that mentioned that it was unknown if a complement of a ribbon 2-knot could have torsion. These groups have presentations that are special cases of what I asked about in the question so I was guessing there was a simple counterexample to my question, but I had no inspiration to find it. Just FYI - the main result of that paper is a well known open problem (are ribbon disk complements aspherical?), so the proof of the main result is flawed... Although I didn't go looking for the error.
Edit : I just wanted to clarify that the elements $g_i$ are arbitrary elements (not necessarily the generators).  So user1729's nice answer answers the question in the case where the $g_i$ are generators, but I am still interested in the question for general $g_i$.  Also,  for those interested in the topological origin of this problem, I actually messed up with the relations that arise in the context of the aforementioned paper.  The groups that arise as ribbon group complements are of the form $F_n / << x_1 = x_2^{g_1}, x_2 = x_3^{g_2},...,x_{n_1} = x_n^{g_n} >>$ where here the $x_i$ are the generators of $F_n$ and again the $g_i$ are arbitrary elements of $F_n$.  I would like to know if these groups can have any torsion as well.  
 A: If the elements $g_i$ are all contained in a single free basis for $F_n$ then the group is torsion free. However, even in this case the answer is non-trivial.
A right-angled Artin group (RAAG), or classically a partially commutative group is a group with generating set $\{x_1, \ldots, x_n\}$ for some $n\in\mathbb{N}$ and relators of the form $[x_i, x_j]$. If the relator set is empty then the group is free, while if every possible relator $[x_i, x_j]$ is present then the group is free abelian (so RAAGS "interpolate" between free and free abelian groups).
Under the assumption that the $g_i$s are in some (fixed) free basis for $F_n$, the group defined by this presentation is clearly a RAAG.
There is an equivalence between RAAGs and graphs: a graph $\Gamma$ defines the RAAG $A_{\Gamma}$ whose generating set consists of the vertex set $V\Gamma$ and the word $[x_i, x_j]$ is a relator if and only if the vertices $x_i, x_j\in V\Gamma$ are connected by an edge. If $\Gamma$ contains no edges then $A_{\Gamma}$ is free, while if $\Gamma$ is a complete graph then $A_{\Gamma}$ is free abelian.
The graph underlying the RAAG here is a straight line (if the second term is $[g_2, g_3]$) or a forest consisting edges of length one (if the second term is $[g_3, g_4]$).
RAAGS are torsion-free. I am unsure of the classical reference for this, but if follows from Theorem 2.6 of Ruth Charney's "An introduction to right-angled Artin groups." Geom. Dedicata (2007) (which you can download from her website here) that every RAAG has a finite $K(\pi_1, 1)$-space, which is the universal cover of its Salvetti complex, and so RAAGs are torsion-free.

I should say a few words about the Salvetti complex: Much of the recent interest in RAAGs is because this is a cubical complex, whose universal cover is a $\mathrm{CAT}(0)$ cube complex, and Haglund and Wise worked some magic with these complexes to understand groups which embed in RAAGS (the key phrase is "special cube complex"). Wise+coauthors then proved that many groups virtually embed into RAAGS (limit groups, one-relator groups with torsion, small cancellation groups, etc.). In the same breath, Wise proved a special case of Thurston's virtually fibered conjecture; Ian Agol then build on Wise's work to prove the virtual Haken conjecture which, when combined with Wise's result, resolved the virtual fibering conjecture. These were two of the last problems in Thurston's program on $3$-manifolds.
