$\newcommand\al\alpha\newcommand\de\delta$The answer is yes. Indeed, take any real $\de>0$ and let $$g(z):=\int_\de^z\frac{dy}{\sqrt{k(\de^{1-\al}-y^{1-\al})}}$$ for real $z\ge\de$, where $k:=\frac2{\al-1}$. Then $g$ is a continuous function strictly increasing from $0$ to $\infty$ on the interval $[\de,\infty)$, and $g$ is smooth on the interval $(\de,\infty)$, whereas $g'(\de+)=\infty$.
For real $t$, let $f(t):=g^{-1}(|t|)$. Then $f$ is a solution to your ODE with $\inf f=f(0)=\de$ and $|f'|=\sqrt{k(\de^{1-\al}-f^{1-\al})}\le\sqrt{k\de^{1-\al}}$, so that $f$ is Lipschitz and hence Hölder.
Detail: On the interval $[0,\infty)$, we have $f'=\sqrt{k(\de^{1-\al}-f^{1-\al})}$ and hence $$f''=\frac{-k(1-\al)f^{-\al}f'}{2\sqrt{k(\de^{1-\al}-f^{1-\al})}} =f^{-\al},$$ so that $f$ satisfies your ODE on the interval $[0,\infty)$. Similarly, $f$ satisfies your ODE on the interval $(-\infty,0]$.