$\newcommand\al\alpha\newcommand\de\delta\newcommand\R{\Bbb R}\newcommand\B{\mathrm B}$The answer is yes. Indeed, take any real $\de>0$ and let 
$$g(z):=\int_\de^z\frac{dy}{\sqrt{k(\de^{1-\al}-y^{1-\al})}}
\Big[=\frac{\de ^{(\al +1)/2} }
{\sqrt{2} \sqrt{\al -1}}\, 
\B_{1-(z/\de)^{1-\al}}\Big(\frac{1}{2},\frac{1}{1-\al }\Big)\Big]$$
for real $z\ge\de$, where $k:=\frac2{\al-1}$ and $\B$ is the [incomplete beta function][1]. Then $g\colon[\de,\infty)\to\R$ 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 $(-\infty,0]$.


  [1]: https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function