Does Gleason's Theorem Imply Born's Rule?

I think I have understood your question now (and I deleted my previous answer since it actually referred to the wrong question). Let me try to summarize.

On the one hand we have a wavefunction $\psi$ in the Hilbert space $L^2(\mathbb R)$ for a given quantum system $S$ and we know that $\psi$ determines the state of $S$ in some (unspecified) sense: it can be used to extract transition probabilities and probabilities of outcomes when measuring observables.

($\psi$ could arise from some analogy optics - mechanics and can have some meaning different from that in Copenaghen intepretation, e.g. a Bohmian wave.)

On the other hand we know, from Gleason's theorem, that an (extremal to stick to the simplest case) probability measure associated to $S$ can be viewed as a a wavefunction $\phi \in L^2(\mathbb R)$ and Born's rule can be now safely used to compute the various probabilities of outcomes.

You would like to understand if necessarily $\psi=\phi$ up to phases as a consequence of Gleason's theorem.

Without further requirements on the procedure to extract transition probabilities (you only say that transitions probabilities can be extracted from $\psi$ with some unspecified procedure) it is not possible to conclude that $\psi=\phi$ up to phases in spite of Gleason's theorem.

We can only conclude that there must be an injective map $$F : L^2(\mathbb R) \ni \psi \to [\phi_\psi] \in L^2(\mathbb R)/\sim$$ where $[\cdot]$ denotes the equivalence class of unit vectors up to phases.

A trivial example of $F$ is $$\phi_\psi := \frac{1}{||\psi+ \chi||}(\psi + \chi)\quad\mbox{and} \quad \phi_{-\chi} := -\chi$$ where $\chi$ is a given (universal) unit vector.

This map is evidently non-physical since there is no reasonable way to fix $\chi$ and, assuming this form of $F$, some argument based on homogeneity of physical space would rule out $\chi$. However much more complicated functions $F$ (not affine nor linear) can be proposed and in the absence of further physical requirements on the correspondence $\psi$-$\phi_\psi$ (e.g. one may assume that some superposition principle is preserved by this correspondence) or on the way to extract probabilities from $\psi$, Gleason's theorem alone cannot establish the form of $F$.

I think Gleason's theorem needs the extra hypothesis of non-contextuality to imply the Born rule. One could, in principle, introduce other measures of probability, but then one violates non-contextuality. See this paper for instance.