This is not really an answer, but rather a meta-answer as to why there exist many conventions in the first place.
The symmetric monoidal category $\mathit{sVect}$ of super-vector spaces has an interesting autoequivalence $J$. The symmetric monoidal functor $J:\mathit{sVect}\to \mathit{sVect}$ is the identity at the level of objects and at the level of morphisms. But the coherence $J(V \otimes W) \xrightarrow{\cong} J(V) \otimes J(W)$ is non-trivial. It is given by $-1$ on $V_{odd} \otimes W_{odd}$ and $+1$ on the rest.
The image of $\mathit{Cliff}(V,q)$ under $J$ is $\mathit{Cliff}(V,-q)$. So anything that you do with one convention can equally well be done with the other convention.
Over the complex numbers, $J$ is equivalent to the identity functor. The symmetric monoidal natural transformation $J\Rightarrow Id$ that exhibits the equivalence acts as $i$ on the odd part and as $1$ on the even part of any super-vector space.
Over the reals, $J$ is not equivalent to the identity functor, as can be seen from the fact that $\mathit{Cliff}(\mathbb R,|\cdot|^2)\not\simeq\mathit{Cliff}(\mathbb R,-|\cdot|^2)$.
One last technical comment: Over $\mathbb C$, the action of $\mathbb Z/2$ on $\mathit{sVect}$ is still non-trivial, despite the fact that $J$ is equivalent to the identity functor. A trivialization of the action isn't just an equivalence $\alpha:J\cong Id$. For such an equivalence to trivialize the action, it would need to satisfy the further coherence $\alpha\circ \alpha = 1$... which can never be satisfied. (To trivialize the action of a group $G$, one needs to trivialize the actions of each $g\in G$ in such a way that the trivializations of $g,h\in G$ compose to the trivialization of $gh$.)
Now, as far as practical things are concerned, I would recommend minimizing the number of minus signs that you end up writing down.