From the iterative inequality (assuming $1-c\gamma_k^\alpha >0$ as said)
$$\gamma_{k+1}^{-\alpha}\ge \gamma_k^{-\alpha}\big(1- c\, \gamma_k^\alpha\big)^{-\alpha}\ge \gamma_k^{-\alpha}+\alpha c\, ,
$$
the last inequality just coming from the convexity inequality $(1-x)^{-\alpha}\ge 1+\alpha x $, for $0 < x < 1$.

Therefore $\gamma_k^{-\alpha}\ge \gamma_0^{-\alpha}+ k\alpha c\ge(k+1) \alpha c$, and your polynomial bound follows with $C:=(\alpha c ) ^{-1/\alpha}$.

$$\bullet$$

As to the other question, there is no uniform constant for all $0<\alpha<1$, and the reason is that the above constant is indeed optimal. Precisely, for any iteration $\gamma_{k+1}= \gamma_k(1- c \gamma_k^\alpha)$ with $1-c\gamma_0^\alpha >0$ we have
$$\gamma_k=Ck^{-1/\alpha}\big(1+o(1)\big)\, ,\quad\mathrm{for}\, k\to\infty \, .$$

Indeed, since $\gamma_k\to 0$, we have
$$
\gamma_{k+1}^{-\alpha}= \gamma_k^{-\alpha}\big(1- c\, \gamma_k^\alpha\big)^{-\alpha}= \gamma_k^{-\alpha}+\alpha c+o(1)\, ,\quad\mathrm{for}\, k\to\infty \, ,
$$
because $(1-x)^{-\alpha}=1+\alpha x +o(x)$ for $x\to0$. Thus $\gamma_{k^{-\alpha}}=k\alpha c+o(k)=k\alpha c\big(1+o(1)\big)$, and the above asymptotic follows.