If L_1 and L_2 are two line bundles on a manifold $M$ that differ by torsion, then their Chern characters
$$ch(L_1) = 1 + c_1(L_1) + \frac{1}{2}c_1(L_1)^2 + \cdots$$
$$ch(L_2) = 1 + c_1(L_2) + \frac{1}{2}c_1(L_2)^2 + \cdots$$
agree, if only because the right-hand sides of these formulas are taking place in $H^*(M;\mathbb{Q})$ and there $c_1(L_1) = c_1(L_2)$. If $L_1$ and $L_2$ are prequantizations of a symplectic structure on M, and if we can give M a compatible complex structure, then we have Q(L_1) = Q(L_2) by the Riemann-Roch formula
$$Q(L_1) = \int ch(L_1) td(M) = \int ch(L_2) td(M) = Q(L_2)$$

We do not get a finer equation than $Q(L_1) = Q(L_2)$. Below the line is an example of $M$, $L_1$, and $L_2$ for which $H^*(M;L_1)$ and $H^*(M;L_2)$ are not isomorphic, that I thought of first. I'll integrate these answers later.

I had written an answer with a bogus example of a Kahler manifold with two prequantum line bundles $L_1$ and $L_2$ for which $Q(L_1)$ was different from $Q(L_2)$. Here's my attempt to repair it--the best I can do is a symplectic manifold with two prequantum line bundles $L_1$ and $L_2$ for which $H^0(L_1) \neq H^0(L_2)$. But there is definitely some higher cohomology that might cancel this difference when you compute $Q$. I don't have a guess for what happens.

I am using "prequantum line bundle" to mean a line bundle whose real chern class (times $2\pi i$?) is equal to the class of the symplectic form. The premise of my example is that if we have two such line bundles $L_1$ and $L_2$ whose integral Chern classes differ by $n$-torsion, then $L_1$ and $L_2$ will become isomorphic over some $\mathbb{Z}/n$-cover $Y$ of $M$. (One can see this by thinking about the bundle-theoretic meaning of the Bockstein map $H^1(M;\mathbb{Z}/n) \to H^2(M;\mathbb{Z})$ and the long exact sequence it fits into.).

We can organize the data this way: $f:Y \to M$ is a $\mathbb{Z}/n$-bundle over $M$ with a $\mathbb{Z}/n$-equivariant prequantum line bundle $L$, and $L_1$ and $L_2$ are built from $L$ by the formula
$$H^0(U;L_i) = \alpha_i\text{-eigenspace of }H^0(f^{-1}(U);L)$$
where $\alpha_1$ and $\alpha_2$ are two different characters of $\mathbb{Z}/n$.

So to find such an $M$, we should find a prequantized symplectic manifold $Y$ equipped with a free $\mathbb{Z}/n$-action, such that $\mathbb{Z}/n$ does not act so uniformly on $H^0(Y;L)$ (i.e. so that the different eigenspaces of a generator for $\mathbb{Z}/n$ have different dimensions). The only source of prequantized symplectic manifolds available to me are projective varieties, and the only source of free $\mathbb{Z}/n$-actions are those on projective hypersurfaces.

So, let $Y$ be the degree 5 hypersurface in $\mathbb{P}^3$ given by the equation $x^5 + y^5 + z^5 + w^5 = 0$. If $\eta$ is a 5th root of unity, then the $\mathbb{Z}/5$-action given by $x \mapsto x$, $y \mapsto \eta y$, $z \mapsto \eta^2 z$ and $w \mapsto \eta^3 w$ is free (to check the absence of fixed points, it's important that the eigenvalues of the different letters are distinct). If $L$ is the restriction of $O(1)$ on $P^3$ to $Y$ then $H^0(Y;L) = H^0(P^3;O(1))$, which is the vector space spanned by $x,y,z,$ and $w$, and the generator of $\mathbb{Z}/5$ acts by $x \mapsto x$, $y \mapsto \eta^{-1} y$, $z \mapsto \eta^{-2} z$, and $w \mapsto \eta^{-3}w$.

The conclusion is that if $L_1$ is the line bundle on $M = Y/(\mathbb{Z}/5)$ corresponding to the eigenvalue 1 (or $\eta^{-1}$ or $\eta^{-2}$ or $\eta^{-3}$) then $H^0(M;L_1)$ is one-dimensional, and if $L_2$ is the line bundle on $M$ corresponding to the eigenvalue $\eta^{-4} = \eta$ then $H^0(M;L_2)$ is zero-dimensional.

More generally one can take (degree n hypersurface in $\mathbb{P}^r$)/(action of $\mathbb{Z}/n$), but to make the action free one needs $n > r+1$ (and maybe $n$ prime). Taking large $n$ like this creates cohomology in $H^{r-1}(M;L_1)$ and $H^{r-1}(M;L_2)$.