Assume $g,f\colon A\subset\mathbb{R}^M\rightarrow\mathbb{S}^2$ are two bijective functions defined on the set $A$. Now assume a constraint $C$: $\forall x,y\in A, \exists R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$. 
Does $C$ imply $f=\pm g$?


[Update]: The answer is "yes" when $C$ is modified to $\forall x,y\in A, \forall R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$. 



Uniqueness of function with range $\mathbb{S}^2$ under a constraint
You are correct, if you change $C$ so $R$ is $\forall$-quantified, then $C$ implies $f = \pm g$.


Pairs $\{v, -v\}$ have distinct stabilizers under $SO(3)$ so setting $x = y$ we see $f = \pm g$ pointwise. But then if $f(x) = g(x)$ and $f(y) = -g(y)$, picking any $R$ such that $Rf(x) = f(y)$, if we were to have $Rg(x) = g(y)$ then we would have
$$ Rf(x) = Rg(x) = g(y) = -f(y) = -Rf(x), $$
a contradiction. Of course neither the choice of domain nor the fact the dimension of the sphere is $2$ matters.


(Your argument is probably correct as well, but I had already written this when I realized $\cdot$ means dot product.)


