I'm a bit late to the party, but here's what I suspect the answer should look like. Forgive me for working somewhat to very heuristically throughout.

To star things off, here's a silly question: why is cohomology $\mathbb{Z}$-graded? Starting from a spectrum $E$ and a space $X$ we can consider the set $[X, E]$ of homotopy classes of maps from the suspension spectrum of $X$ to $E$. This only gives the zeroth part $E^0(X)$ of the $E$-cohomology of $X$. To get more interesting information one thing we can do is smash $E$ with invertible spectra. These are precisely the shifts $S^n, n \in \mathbb{Z}$ of the sphere spectrum, and

$$X \mapsto [X, S^n \wedge E]$$

recovers $E$-cohomology as ordinarily understood. So we should think of $\mathbb{Z}$ here as being the Picard group $\text{Pic}(\text{Sp})$ of the symmetric monoidal $\infty$-category of spectra. That is:

Cohomology in general is graded by the Picard group of invertible spectra.

In more complicated situations we get more interesting Picard groups that we can grade by. For example, $\infty$-local systems of spectra over a space $X$ describe cohomology theories for spaces over $X$, and these cohomology theories are therefore graded by the Picard group $\text{Pic}(\text{Loc}_{\text{Sp}}(X))$. The fiberwise suspension spectrum of a spherical fibration over $X$ is in particular an example of an element of this Picard group. This should be relevant to André Henriques' observation in the comments.

One of the more complicated situations we could be in is the following. Let $R$ be an $E_{\infty}$ ring spectrum and let $E$ be an $R$-module spectrum. Rather than settling for smashing $E$ with invertible spectra, we can now smash $E$ with invertible $R$-module spectra over $R$ to get more interesting information of the form

$$X \mapsto [X, A \wedge_R E]$$

where $A$ is invertible with respect to $\wedge_R$; that is, $E$-cohomology acquires a grading by $\text{Pic}(R\text{-ModSp})$, which I will abbreviate to $\text{Pic}(R)$. In particular, letting $R = E = KO$:

Real K-theory is graded by the Picard group of invertible $KO$-module spectra.

The punchline should now be that there is a natural group homomorphism from the super Brauer group of $\mathbb{R}$ to the Picard group of $KO$; possibly this is even an isomorphism. The construction should go something like this:

- Start with a finite-dimensional superalgebra $A$ over $\mathbb{R}$.
- Construct the category of finite-dimensional (projective?) right $A$-supermodules, which is symmetric monoidal with respect to direct sum and enriched over supervector spaces. Topologize its homs as finite-dimensional real vector spaces.
- Take the core of the above, which is a symmetric monoidal topological groupoid, and hence which presents a symmetric monoidal $\infty$-groupoid.
- Take the group completion of the above, which is a grouplike $E_{\infty}$ $\infty$-groupoid, or equivalently an infinite loop space.

But this isn't quite right, because we only get a connective spectrum this way. I don't know how to fix this. Maybe we should take chain complexes or something.

Anyway, let's pretend I did the right thing, and let's call the resulting spectrum $K(A)$. Let's also pretend that $K(\mathbb{R})$ ended up being $KO$. The categories I constructed above are all tensored over supervector spaces, or equivalently are all module categories over supervector spaces, so the resulting spectra should all be $KO$-module spectra. Moreover, taking tensor products should give natural equivalences

$$K(A) \wedge_{K(\mathbb{R})} K(B) \cong K(A \otimes_{\mathbb{R}} B)$$

and since $K(A)$ only depends on the category of $A$-modules, any Morita-invertible $A$ should give an invertible $K(\mathbb{R})$-module spectrum $K(A)$ in a way that respects products.

So real K-theory has two gradings, one by $\text{Pic}(S)$ ($S$ the sphere spectrum) and one by $\text{Pic}(KO)$. The content of the Clifford-algebraic proofs of Bott periodicity should be that these gradings collapse into a single grading, and in particular that all of the $K(A)$ end up being shifts of $K(\mathbb{R})$ as $K(\mathbb{R})$-module spectra.

Of course we can replace $\mathbb{R}$ by $\mathbb{C}$ throughout. Actually it seems like we really ought to replace $\mathbb{R}$ by an arbitrary ring spectrum $R$ throughout, and try to relate the Brauer group of $R$ and the Picard group of $K(R)$; this might make it easier to figure out the right statements.