Let $G$ be the group with presentation $$G = \langle \sigma_1 ,\sigma _2 , \sigma _3\, |\, \sigma _1^2=\sigma _2^2=\sigma _3^2=(\sigma _1\sigma _2)^p =(\sigma _2\sigma _3)^q =(\sigma _3\sigma _1)^r =1\rangle .$$ I want to have a presentation for the subgroup $H$ generated by the elements $\tau _1=\sigma _1\sigma _2,\, \tau _2=\sigma _2\sigma _3,\ \tau _3=\sigma _3\sigma _1$.
Is it enough to conjecture the presentation $$ H=\langle \tau _1,\tau _2 ,\tau _3\, |\, \tau _1^p=\tau _2^q=\tau _3^r=\tau _1\tau _2\tau _3 =1 \rangle$$ and argue that the relations in the presentation of $G$ clearly imply the ones conjectured for $H$ and that the relations of $H$ imply the relations in the presentation of $G$.
No, this is not sufficient, as there may be "hidden" relations. In your example, $H$ does indeed have the given presentation, and it is presumably possible to show this "by hand" using properties of the group $G$, but it is not "clearly" true.
Alternatively, you can use the Reidemeister-Schreier method. This is a process to find a presentation of a given subgroup of a group, and it works nicely here because $H$ has finite index in $G$. There are two approaches to the method: the classical combinatorial/algebraic version as covered in Section 2.3 of the book Combinatorial group theory by Magnus, Karrass and Solitar, or in Section II.4 of Lyndon and Schupp's book of the same name, and the topological version using coverings of presentation complexes (see here). Personally, I favour the topological version, as I view it as a "visualisation" of the algebraic version.
The process can also be done on a computer (see here), although the computer will want $p,q,r$ to be fixed.