Is it possible to continuously select a probability distribution over fixed points in Brouwer's fixed point theorem?

There can be no such function even in the category of smooth functions.

Here an example for functions $f:[0,1]\to [0,1]$ with $f(0)=\epsilon$ and $f(1)=1-\epsilon$:

(1) Let $\operatorname{id}:x\mapsto x$ and choose a smooth function $f_0$ with $f_0=\operatorname{id}$ on $[\frac{1}{3},\frac{2}{3}]$, $f_0>\operatorname{id}$ on $[0,\frac{1}{3})$ and $f_0<\operatorname{id}$ $(\frac{2}{3},1]$.

(2) There is a sequence of smooth functions $f_n$ converging to $f_0$ with $f_n(\frac{1}{3})=\frac{1}{3}$, $f_n>\operatorname{id}$ on $[0,\frac{1}{3})$ and $f_n<\operatorname{id}$ on $(\frac{1}{3},1]$.

(3) There is a sequence of smooth functions $g_n$ converging to $f_0$ with $g_n(\frac{2}{3})=\frac{2}{3}$, $g_n>\operatorname{id}$ on $(0,\frac{2}{3})$ and $g_n<\operatorname{id}$ on $(\frac{2}{3},1]$.

Both sequences have each a unique fixed point equal to $\frac{1}{3}$ and resp. $\frac{2}{3}$. Thus there are exactly two distinct probability measures associated to the fixed point sets of $f_n$ and resp. $g_n$. Obviously the limits of those probability measures do not agree. In particular, any choice of $fix$ cannot be continuous at $f_0$.