I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose $\phi\in C^\infty(\mathbb{R})$, $\phi\ge0$ and $\int_{\mathbb{R}}\phi(x)dx=1$, and let $\phi_R(x)=R\phi(Rx)$. Define
$$ f=\phi_R\star\eta,\quad g=\eta-f.$$
Then it is easy to see that
$$ \|f\|{C^n}=O(R^{n-\alpha}),\quad \|g\||f\|_{C^n}=O(R^{n-\alpha}),\quad \infty=O(R^{-\alpha}),$$ |g\|_\infty=O(R^{-\alpha}),$$
but this is not what you are asking for.
My feeling is that the constant $C$ must show some dependence on $n$.
In response to your last comment, let me prove the estimate on $\|f\|_{C^n}$. We have
$$f^{(n)}=(\phi_R)^{(n)}\star\eta=R^n(\phi^{(n)})_R\star\eta.$$
Since $(\phi^{(n)})_R$ has mean zero, for any $x\in\mathbb{R}$:
$$ |f^{(n)}(x)|\le R^n\int_{\mathbb{R}}|\phi^{(n)}(y)||\eta(x-\frac{y}{R})-\eta(x)|dy\le HR^{n-\alpha}\int_{\mathbb{R}}|\phi^{(n)}(y)||y|^\alpha dy,$$
where $H$ is $\eta$'s Hölder constant.
