It seems there aren't any counterexamples, even if $f$ is homogeneous but not even.

If there is some counterexample $f$ to the question, and letting $X=\mathbb{R}^3\setminus\{x_1=x_2=x_3\}$, we can consider the map $F:SO(3)\to X$, given by $F(u,v,w)=(f(u),f(v),f(w))$ (we will sometimes represent elements $r\in SO(3)$ by $(r(e_1),r(e_2),r(e_3))$, $(e_i)_{i=1}^3$ being the usual basis of $\mathbb{R}^3$).

As $\pi_1(SO(3))\cong\mathbb{Z}_2$ and $\pi_1(X)\cong\mathbb{Z}$, we will have $F_*(\gamma)=0\;\forall\gamma\in\pi_1(SO(3))$.

Now consider the loop $\gamma:t\mapsto\gamma_t=r_{2\pi t}(v)$ given by rotations of angle $2\pi t$ around $v:=(1,1,1)$. Note that $\gamma_{\frac{1}{3}}(e_1)=e_2,\gamma_{\frac{1}{3}}(e_2)=e_3$ and $\gamma_{\frac{1}{3}}(e_3)=e_1$. So if $\gamma_t=(u,v,w)$, then $\gamma_{t+\frac{1}{3}}=(\gamma_{t+\frac{1}{3}}(e_1),\gamma_{t+\frac{1}{3}}(e_2),\gamma_{t+\frac{1}{3}}(e_3))=(\gamma_{t}(e_2),\gamma_{t}(e_3),\gamma_{t}(e_1))=(v,w,u)$.

So the path $\alpha=F_*\gamma\in\pi_1(X)$ satisfies that if $\alpha(t)=(x_1,x_2,x_3)$, then $\alpha(t+\frac{1}{3})=(x_2,x_3,x_1)$. From this we will deduce that $\alpha$ is not trivial, which is a contradiction. To deduce that $\alpha$ is not trivial, we can first deformation retract $X$ to a circumference perpendicular to the line $\{x_1=x_2=x_3\}$, so that identifying this circumference with $\mathbb{S}^1$, the rotation $(x_1,x_2,x_3)\mapsto(x_2,x_3,x_1)$ becomes the rotation $r_{\frac{2\pi}{3}}$ of $\mathbb{S}^1$.

So we have to prove that any loop $\beta:[0,1]\to\mathbb{S}^1$ satisfying $\beta(t+\frac{1}{3})=e^{\frac{2\pi i}{3}}\beta(t)$ is not trivial. This is true because as $\beta(\frac{1}{3})=e^{\frac{2\pi i}{3}}$, the path $\beta|_{[0,\frac{1}{3}]}$ rotates a total angle of $2\pi(n+\frac{1}{3})$ for some $n\in\mathbb{Z}$, so the winding number of $\beta$ is $3(n+\frac{1}{3})\neq0$.