I'm sorry for not reacting for a long time. I have to admit my question was a bit vague, but I have found the solution, I was looking for, now. There is a complete isomorphism-invariant $\theta(\Phi)\in\mathbb{Z}_2$ of the irreducible, complex, finite dimensional representations $\Phi$ of the Clifford-Algebra $Cl(V)$ of an odd-dimensional, complex, non-degenerate quadratic space $(V,q)$ defined as follows. Assume we have chosen a preferred orientation in $V$ and consider an oriented orthonormal basis $b_1\dots,b_n$ of $V$. Then we call the element $\eta=i^{n(n-1)/2}b_1\cdots b_n\in Cl(V)$ the volume element of $Cl(V)$, and the Clifford Relations imply that $\eta$ does not depend on the chosen basis. The factor is chosen such that $\eta^2=1$, hence $\Gamma=\Gamma(\Phi):=\Phi(\eta)$ also satisfies $\Gamma^2=1$ and decomposes the representation space $S$ of $\Phi$ into $S=S_+\oplus S_-$ with $S_\pm:=\ker \Gamma\mp 1$. Since $n$ is odd, $\eta$ lies in the center of $Cl(V)$ and thus the $S_\pm$ are $Cl(V)$-invariant subspaces of $S$. By irreducibility, we thus have either $S_+=S$ or $S_-=S$, i.e. $\Gamma=\theta(\Phi)\cdot\operatorname{id}_S$ for some $\theta(\Phi)\in\mathbb{Z}_2$. One then easily computes that $\theta(\Phi)$ only depends on the isomorphism class of $\Phi$ and that $\theta(\Phi\circ\chi)=-\theta(\Phi)$, where $\chi\in\operatorname{Aut}(Cl(V))$ is the canonical involution on $Cl(V)$, defining the $\mathbb{Z}_2$-grading. Therefore $\Phi\not\cong\Phi\circ\chi$ and by the structure theorem, these are the only irreducible representations (upto isomorphism).
To summarize, two irreducible (complex, finite-dimensional) representations $\Phi,\Psi$ of $Cl(V)$ are isomorphic if and only if both $\Phi(\eta)$ and $\Psi(\eta)$ both act as either plus or minus the identity on the corresponding representation space.
Cheers, Robert