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
$$f=\phi_R\star\eta,\quad g=\eta-f.$$
$$\|f\|{C^n}=O(R^{n-\alpha}),\quad \|g\||f\|_{C^n}=O(R^{n-\alpha}),\quad \infty=O(R^{-\alpha}),$$ |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,$$where H is \eta's Hölder constant. 1 I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose \phi\in C^\infty(\mathbb{R}), \phi\ge0 and \int_{\mathbb{R}}\phi(x)dx=1, and let \phi_R(x)=R\phi(Rx). Define$$ 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\|\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 $n$.