I believe I have a valid counterexample for $n > 1$, unfortunately I don't have the expertise to be certain. I would glad if someone would expand or refute the following.
Take $n=2$ to start. Now we are going to create a pair of functions $f_1(x),f_2(x)$ with $g(x)=(f_1,f_2)$ which don't satisfy the property. First take $f=f_1'=f_2'$ to be a simple bipolar function such that $f$ is $0$ on a great circle and has a maximum at $x_m = (1,1)$ and minimum at $(-1,-1)$ s.t. $g(x_m) = M < 1$.
Now we modify $f_1'$ into $f_1$ to have a continuous bump around $x_m$ of radius $r_1$ and value $1$, that is, $g(x) \approx (f_1',*)$ when $|x_m \cdot x| \le 1-h(r_1)$ and $g(x) \approx 1$ when $|x_m \cdot x| \gt 1-h(r_1)$, and accordingly we modify $f_2$ to have a bump around $x_m$ of radius $r_2$ and value $1$, that is, $g(x) \approx (*,f_2')$ when $|x_m \cdot x| \le 1-h(r_2)$ and $g(x) \approx 1$ when $|x_m \cdot x| \gt 1-h(r_2)$. So that we have $g=(f_1,f_2)$.
Now if $h(r_1) \ll h(r_2) \ll 1$, then $f_1$ assumes almost all the values in $[M,1]$ near a circle of radius $r_1$, while $f_2$ assumes almost all values in $[M,1]$ near a circle of radius $r_2$. Clearly, those concentric circles don't intercept. We conclude there exists $y \in [M,1]$ such that $g(x)=(y,y)$ has no solution.
I could generalize this, if I'm not missing something.