$\newcommand\al\alpha\newcommand\be\beta$The answer is yes.
Indeed, using the inequalities $u^\be\le\max(1,u)\le1+u$ for $u=f(s)\ge0$, we see that for $t\in[0,1]$ the inequality $$f(t) \le \al+ \int_0^t (t-s)^{-1/2} [f(s) + f(s)^\be] \,ds$$ implies $$f(t) \le \al+2t^{1/2}+2 \int_0^t (t-s)^{-1/2} f(s) \,ds.$$
It remains to use Theorem 2.2, which yields $$f(t) \le (\al+2t^{1/2})E_{1/2}(2\sqrt\pi\,t^{1/2}) \le E_{1/2}(2\sqrt\pi)(\al+2),$$ where $E_b$ is the Mittag--Leffler function, which is an entire function given by series $E_b(z):=\sum_{k=0}^\infty\frac{z^k}{\Gamma(bk+1)}$ for complex $z$.