It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbitrary, representation) does not. Here two extensions $K$ and $K'$ of a field $k$ are stably isomorphic if for some $m$ and $n$ $K(x_1,\ldots,x_m)$ and $K'(x'_1,\ldots,x'_n)$ are isomorphic as $k$-extensions.
Addendum: The result is well-known but I cannot at the moment come up with a reference so instead I give the proof: Let $V$ be a faithful $G$-representation and $U$ the non-empty Zariski open subset where $G$ acts freely. Then $k(V)^G$ is the fraction field of $U/G$. If now, $V'$ is another faithful representation with open subset $U'$ we have that $U\times V'$ has a free $G$-action with a linear action on the second factor. Hence $U\times V'/G$ is a vector bundle over $U/G$ (by descent theory) and in particular its fraction field is stably isomorphic to that of $U/G$. However, $U\times V'/G$ is birational to $U\times U'/G$ which in turn is birational to $V\times U'/G$. The fraction field of the latter is for the same reason stably isomorphic to that of $U'/G$.