My humble point of view is that the Witten-Reshetikin-Turaev invariant (at least for $G=SU(2)$ or $SO(3)$) at the root $\xi$ is (by definition) a rational function on $\xi$ which looks very much like a polynomial with integer coefficients. (Note that the function depends on $\xi$.)
When $M=S^3$, the rational function is indeed a polynomial with integer coefficients that does not depend on $\xi$: that's the (colored) Jones polynomial. So the Witten-Reshetikin-Turaev invariant of $L\subset S^3$ lies in $\mathbb Z[\xi]$ in that case. And it is natural to ask whether this holds for general $M$.
When $M\neq S^3$ the rational function is no longer a polynomial and moreover it depends of $\xi$ (but this dependence is not an issue concerning integrality). The numerator of this rational function is an arbitrary element in $\mathbb Z[\xi]$, but the denominator is not arbitrary: it is just a product of quantum integers $$[n] = \frac{\xi^n - \xi^{-n}}{\xi - \xi^{-1}} = \xi^{n-1} + \xi^{n-3} + \ldots + \xi^{-n+1}.$$ Hence the invariant is "almost" a polynomial with integer coefficients: it is an element in $\mathbb Z[\xi]$ divided by a monomial like $[2]^3[4]^2[7]$. (I am probably ignoring some renormalisation factor.)
As far as I remember, when $\xi$ is a $4r$-th root of unity and $r$ is prime, then every $[n]$ is actually invertible in $\mathbb Z[\xi]$, and hence the invariant indeed lies in $\mathbb Z[\xi]$.