First of all, $X$ being non-lattice is not enough for $X_t$ being absolutely continuous. A simple counter-example is $$X_t = \sum_{n=1}^\infty \frac{N_t^{(n)}}{n!} \, ,$$ where $N_t^{(n)}$ are independent Poisson processes. If $X$ is non-lattice and not a compound Poisson process, then the distribution of $X_t$ is continuous. However, the definition of the quantile $Q_t(p)$ does not depend on continuity $X_t$; what is needed here is that the support of $Q_t(p)$ is an interval. This is the case when $X$ has a Gaussian component or when the closed semigroup generated by the support of Lévy measure of $X$ is an interval. (A detailed discussion of distributional properties of a Lévy process can be found, for example, in Sato's book *Lévy processes and infinitely divisible distributions*.)

It is quite simple to see that $Q_t(\tfrac12)$ need not be linear. Consider first the usual Poisson process $X_t$. Then $Q_t(\tfrac12) = 0$ for $t < \log 2$, and $Q_t(\tfrac12) = 1$ when $\log 2 < t < T$, where $T$ satisfies $e^{-T} (1 + T) = \tfrac12$. In particular, $Q_t$ is not linear in this case.

If one insists on a Lévy process $X_t$ with a continuous distribution, then it is enough to add an "epsilon" of Brownian motion to the above example. This is intuitively clear, I hope, but a detailed argument would require a somewhat technical estimate.