EDIT: After some clarification from the questioner with respect to the $Q$-Gorenstein definition, let me point out that additional hypotheses are needed I think.
I don't think there's any hope that this can be true without additional hypotheses. I'll start by claiming that I can pick $\mathcal{E, F}$ such that $\text{Ext}^i(\mathcal{E,F}) \neq 0$ for some $i > n = \dim X$, even if $X$ is $Q$-Gorenstein.
For example, since $X$ is not Gorenstein, $O_X$ does not have finite injective dimension, and fix a pure $\mathcal{E'}$ such that $\mathcal{E}xt^i(\mathcal{E'}, O_X) \neq 0$ for some $i > n$ (we can find such modules that are pure for $i - 1 > n$ , write any such module with that nonvanishing in a short exact sequence as a quotient by a free module, the kernel should do the job and will be pure). By setting $\mathcal{E} = \mathcal{E}' \otimes L^{-m}$ for some ample $L$ and $m \gg 0$ and setting $\mathcal{F} = O_X$, we also obtain $\text{Ext}^i(\mathcal{E,F}) \neq 0$ for some $i > n$ by Serre vanishing.
But $\text{Ext}^{n-i}(\mathcal{F}, \mathcal{E}\otimes \omega_X) = 0$ for $i > n$ for obvious reasons.
ORIGINAL Answer:
First let me clarify something, when you say that $\mathcal{E}$ and $\mathcal{F}$ are pure, you mean sheaves of pure dimension? If so, then you are saying that every subsheaf of $\mathcal{E, F}$ is of the same dimension. In particular, line bundles on a variety $X$ are always pure?
Now, when you say $\omega_X$ is a dualizing sheaf, this actually doesn't really make sense unless $X$ is Cohen-Macaulay, because otherwise there is no dualizing sheaf, but there is a dualizing complex. If $X$ is normal, by $\omega_X$ you probably mean the canonical sheaf $i_* \Omega_{X^{\text{reg}}/k}$ where $i : X^{\text{reg}} \to X$ is the inclusion of the regular locus.
Definition: Suppose that $K_X$ is a divisor such that $O_X(K_X) \cong \omega_X$ then we say that $X$ is $Q$-Gorenstein if $nK_X$ is a Cartier divisor for some $n > 0$.
In this case, we now run into a problem with this very unfortunately terminology.
Fact: When people say the word $Q$-Gorenstein, they do NOT typically require that $X$ is Cohen-Macaulay. This is in start constrast to the term Gorenstein which includes in its definition the fact that $X$ is Cohen-Macaulay (Gorenstein means Cohen-Macaulay + $K_X$ being Cartier).
There are some exceptions to this, but they are rare in algebrabraic geometry.
Now we come to your statement.
Answer: Your statement holds for all $\mathcal{E,F}$ line bundles on a projective variety $X$ if and only if $X$ is Cohen-Macaulay.
You are probably already familiar with the section in Hartshorne on Serre duality.
I will try to briefly explain why they aren't equal. Suppose that $\mathcal{E}$ and $\mathcal{F}$ are line bundles and that $\mathcal{F} = O_X$ and $\mathcal{F}$ is a high power of an ample line bundle.
Then
$$Ext^i(\mathcal{E, F}) = H^i(X, \mathcal{E}^{\vee}), \text{ and } Ext^{n-i}(\mathcal{F}, \mathcal{E} \otimes \omega_X) = H^{n-i}(X, \omega_X \otimes \mathcal{E})^\*$$
In fact, if these two groups are equal in general, then it follows that $X$ is Cohen-Macaulay.
First note that $H^{n-i}(X, \omega_X \otimes \mathcal{E})^\* = 0$ for all $i < n$ by Serre vanishing.
Now we work on the other side. Being Cohen-Macaulay is equivalent to requiring the dualizing complex $\omega_X^{\bullet}$ to be a sheaf. Otherwise, the dualizing complex is a complex unsurprisingly. In general, $X$ is Cohen-Macaulay, if and only if $\omega_X^{\bullet}$ only has cohomology in $-\dim X = -n$ (and has zero cohomology elsewhere).
Now, or a line bundle $L$ that is a high power of an ample line bundle
$$H^i(X, L^{\vee}) = \mathbb{H}^{n-i}(X, L \otimes \omega_X^{\bullet}[-n])$$ (this is a very special case of Grothendieck duality).
However, the right side is simply $H^0(X, L\otimes {\bf h}^{-i}(\omega_X^{\bullet}))$ by analysis of the spectral sequence (here ${\bf h}^j$ simply means take the $j$th cohomology of the complex of sheaves). On the other hand, since $L$ is a high power of an ample line bundle, $H^0(X, L^{\vee} \otimes {\bf h}^{-i}(\omega_X^{\bullet}) \neq 0$ as long as ${\bf h}^{-i}(\omega_X^{\bullet}) \neq 0$. Thus if $\omega_X^{\bullet}$ has cohomology not in $-\dim X$, then there will be some non-vanishing cohomology.
Let me know if this makes sense.
On the plus side: If you are willing to work with Grothendieck duality and not Serre duality, then there are probably statements like the one you want.
Indeed, the point of this is that somehow the tensor functor you are writing there is not the correct functor (you may need the derived version for example).
Best Answer
This works directly when one of $F, G$ is locally free.
I am not sure whether it is true when both are merely assumed to be coherent (e.g. I don't see how to get the map). (In general, even the generalization of Serre duality -- Grothendieck duality -- tells you how to hom out of $\mathbf{R}\Gamma \mathcal{F}$ (or more generally derived push-forward) of a sheaf $\mathcal{F}$ into some complex of abelian groups, and this doesn't seem to tell you about $\mathrm{Ext}$ functors up top, in $X$, though perhaps I'm missing something).See below for the extension without local freeness hypotheses.Namely, there is a map $H^n(X, \omega) \to k$ (the "integration" map*). To get the map $$\mathrm{Ext}^i(F, G) \to \mathrm{Ext}^{n-i}(G, F \otimes \omega)^*$$ (which is natural), we need a pairing $$\mathrm{Ext}^i(F, G) \times \mathrm{Ext}^{n-i}(G, F \otimes \omega) \to k.$$ To do this, we can use the Yoneda product to pair these to $\mathrm{Ext}^n(F, F \otimes \omega)$. If $F$ is locally free, then this naturally maps to $\mathrm{Ext}^n(O_X, F \otimes F^{\vee} \otimes \omega)$, which in turns maps to $H^n(X, \omega)$ (by coevaluation) and thus to $k$. If $G$ is locally free, we can similarly write both sides as $\mathrm{Ext}^i(F \otimes G^{\vee}, \omega)$ and $\mathrm{Ext}^{n-i}(O_X, G^{\vee} \otimes F \otimes \omega)^*$, and we get the pairing and isomorphism just as in Hartshorne.
Now if we fix one of $F, G$, we get a $\delta$-functor in the other. So if the natural transformation is an isomorphism when both are locally free, it is an isomorphism when one is locally free and the other merely coherent (since on a projective scheme, every coherent sheaf has a locally free presentation, and we can use the "finite presentation trick").
*Here the comparison is as follows: on a compact complex manifold $X$ of dimension $n$, if $\omega$ denotes the sheaf of holomorphic $(n,0)$-forms, we have (Dolbeaut isomorphism) $$H^n(X, \omega) = \frac{(n,n)\mathrm{-forms}}{\overline{\partial}\mathrm{-exact\ top forms}}$$ and so we can define the map as integration, legitimately (because a $\overline{\partial}$-exact top form is exact in the usual sense, this is well-defined).
Edit: As above, the obstacle to defining the map was that there was no natural trace morphism $$\mathrm{Ext}^n(F, F \otimes \omega) \to H^n(X, \omega);$$ given one, the same arguments would answer your question for the case of $F, G$ both only assumed coherent. As Donu Arapura observes below, there is a natural way to define the trace. Reason: $F$ can be replaced by a bounded complex of locally frees in the derived category (it is a "perfect" complex) since we are working over a smooth variety (in particular, this means that any locally free resolution can be truncated at a finite stage to still yield a locally free one, by Serre's theorem on the finiteness of global dimension). For a bounded complex of locally frees $K^\bullet$, we can define a map $$\mathrm{Ext}^n(K^\bullet, K^\bullet \otimes \omega) \to H^n(X, \omega)$$ by taking the "partial trace." One can think of the former as consisting of maps $K^\bullet \to K^\bullet \otimes \omega[n]$, or $\mathbf{R}\underline{Hom}(K^\bullet, K^\bullet) \to \omega[n]$. (By the conditions on $K^\bullet$, the derived internal hom is the same as the usual sheaf hom.) Since there is a natural map from $\mathcal{O}_X$ to the derived internal hom (given by the identity), we can define the trace. To show that map you are interested in becomes an isomorphism, we use the same "finite presentation" trick.