It is a basic fact in representation theory of finite groups over complex numbers that the character tables of $Q\_8$ and $D\_8$ are identical. I believe, this implies that the corresponding categories of representations are equivalent (as tensor categories). 


On the other hand, Tannakian Formalism tells us that we can reconstruct a finite group $G$ from its category of representations $\mathbf{Rep}\_G$ together with the natural (foregetfull) fibre functor $F\_G: \mathbf{Rep}\_G\rightarrow \mathbf{Vect}$. Namely, $G$ is canonically isomorphic to the tensor automorphisms of the tensor functor $F\_G$. 


This implies that the fibre functors $F\_{D\_8}$ and $F\_{Q\_8}$ are different. How can one see this explicitly in terms of representations of $D\_8$ and $Q\_8$? 



Tannakian Formalism for the Quaternions and Dihedral Group
Let $V\_D$ and $V\_Q$ be the two dimensional simple representations of $D\_4$ and $Q\_8$ respectively. Let $1\_D$ and $1\_Q$ denote their trivial representations.


Suppose that there is a tensor equivalence between $\mathbf{Rep}(D\_4)$ and $\mathbf{Rep}(Q\_8)$ commuting with the fibre functor to $\mathbf{Vect}\_\mathbb{C}$. This equivalence sends $1\_D$ to $1\_Q$ (as they're the unit object) and sends $V\_D$ to $V\_Q$ (as they're the unique simple of dimension 2).


In particular there is a $\mathbb{C}$-linear isomorphism $g$ from $V\_D$ to $V\_Q$. Consider $$g\otimes g:V\_D\otimes V\_D\to V\_Q\otimes V\_Q.$$
It must send the unique copy of $1\_D$ in $V\_D\otimes V\_D$ to the unique copy of $1\_Q$ in $V\_Q\otimes V\_Q$.


It is easy to see that there is no such $g$. The slickest way I can see to prove this is to note that the flip map $v\otimes w\mapsto w\otimes v$ acts by -1 on $1\_Q$ and by 1 on $1\_D$.


