I guess that Mariano is right:

Assume that $Q$ and $Q'$ are acyclic and mutation-equivalent quivers. Following the result of Keller and Yang, this implies, as Mariano noticed, that their Ginzburg dg algebras, say $A$ and $A'$, are derived equivalent. This equivalence induces an equivalence between the generalised cluster category of $A$ and that of $A'$ ([the generalised cluster category][1] of $A$ is defined as the Verdier quotient of the perfect derived category of $A$ by the bounded derived category of $A$, it is well-defined in the present situation). Since we started with acyclic quivers, the generalised cluster category of $A$ coincides with the [usual cluster category][2] $\mathcal C_Q$ (and the same holds true for $Q'$).

Thus, the cluster categories $\mathcal C_Q$ and $\mathcal C_{Q'}$ are equivalent. Therefore, their Auslander-Reiten quivers are isomorphic. The link between $Q$ and $Q'$ then follows from that fact. More precisely:

The paths algebras of $Q$ and $Q'$ [have the same representation type][3], hence $Q$ is of Dynkin type if and only if so is $Q'$, in which case the Auslander-Reiten quiver of $\mathcal C_Q$ is isomorphic to $\mathbb{Z}Q/\langle\sigma\rangle$ for some automorphism $\sigma$ of $\mathbb{Z}Q$. Since the translation quivers $\mathbb{Z}Q$ and $\mathbb{Z}Q'$ are universal covers of $\mathbb{Z}Q/\langle\sigma\rangle$ and $\mathbb{Z}Q'/\langle\sigma\rangle$, respectively, there is an isomorphism $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ (the covering is understood in the sense of "Covering spaces in representation theory" by Bongartz and Gabriel, Invent. Math. 65 (1982) n°3, 3331--378).

Now assume that neither $Q$ nor $Q'$ is of Dynkin type. Then the Auslander-Reiten quiver of $\mathcal C_Q$ has a [unique connected component with only finitely many $\tau$-orbits][5], it is called the transjective component, and it is isomorphic to $\mathbb{Z}Q$. Therefore $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ in any case.


The isomorphism $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ implies that the paths algebras of $Q$ and $Q'$ have equivalent bounded derived categories and, therefore, that $Q$ and $Q'$ are related by a sequence of reflections (or APR-tilts, see 4.8 in the article "On the derived category of a finite dimensional algebra" of Happel, Coment. Math. Helv., 62 (1987) 339--389).

There should be a shorter arugment, though.

  [1]: http://arxiv.org/abs/0805.1035
  [2]: http://arxiv.org/abs/math/0402054
  [3]: http://arxiv.org/abs/math/0402075
  [4]: http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=bongartz&s5=gabriel&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq
  [5]: http://arxiv.org/abs/math/0402054