Here I will try to record what I understand about the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$. I will suggest that the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$ is a $\mathbb Z/4$, but I will not provide a complete proof. Perhaps someone else will, or will provide a reference, or will point out something I missed.
Any Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb H}$ must complexify to a Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb C}$; the latter are either super matrix algebras or super matrix algebras tensored with (complex) Cliff(1). Conversely, doing the descent is explained above, and it's reasonably clear that if a complex Morita-invertible superalgebra does descend to $\mathrm{SuperVect}_{\mathbb H}$, then its descendent is Morita-invertible.
So now let me try to find some Morita-invertible algebras in $\mathrm{SuperVect}_{\mathbb H}$. A few observations:
We have, of course, the purely even algebras coming from the non-super Brauer group of $\mathbb R$, namely $\mathbb R$ and $\mathbb H$ (the latter is a real form of $\mathrm{Mat}(2)$). But in this quaternionic world, there is a Morita equivalence $\mathbb R \simeq \mathbb H$. Indeed, denote by $\mathbb J$ the purely-odd simple object in $\mathrm{SuperVect}_{\mathbb H}$; then $\mathrm{End}(\mathbb J) = \mathbb H$ by construction, and so $\mathbb J$ furnishes the claimed Morita equivalence. (This is in contrast to the real and complex worlds, where the Picard group — the group of $\otimes$-invertible objects in the category — is a $\mathbb Z/2$ consisting of the even and odd simple objects; here the odd simple does not produce an auto-Morita-equivalence of $\mathbb R$ in the Brauer group, but rather a nontrivial equivalence.)
The Clifford algebras are formed by "quantizing" (graded) symmetric algebras on purely odd vector spaces. By passing to the world of complex super vector spaces with antilinear (anti)involutions, as described above, one can check that $\mathrm{Sym}^2(\mathbb J) \cong \mathbb R^{\oplus 2} \oplus \mathbb J$. Since $\mathbb J$ complexifies to $\mathbb C^{\oplus 2}$, one should hope to quantize this to a "quaternionic" form of $\mathrm{Cliff}(2)$. If I did the check right, there is exactly one "quaternionic" form of $\mathrm{Cliff}(2)$, and it has the property that its even part is the algebra $\mathbb C$. So this feels like a version of $\mathrm{Cliff}(1,-1)$, except a dimension count shows that it is Morita non-trivial.
By a dimension count, $\mathrm{Cliff}(1)$ does not have any quaternionic forms, but $\mathrm{Cliff}(3)$ might. In fact, $\mathrm{Cliff}(3)$ has precisely two quaternionic forms; their even parts are $\mathrm{Mat}(2,\mathbb R)$ and $\mathbb H$ respectively. (Their odd parts have dimension $\mathbb J^{\oplus 2}$.) Again a dimension count verifies that they are not Morita-trivial — of course, this can also be seen by complexifying.
My intuition is that this is it — that these four Morita-classes are all invertible and that there aren't any others. I don't have a strong argument in support of this intuition, and by no means do I claim a proof.
It was suggested to me that "quaternionic K-theory" usually means "symplectic K-theory" KSP. That's an 8-periodic theory which does not have products — it is simply a shift by 4 of KO theory. If there is a "K-theory" associated with $\mathrm{SuperVect}_{\mathbb H}$, probably it does have products, and my intuition is that it is 4-periodic. Perhaps it is something like "4-periodicitized KO theory".
