From the iterative inequality (assuming $1-c\gamma_k^\alpha >0$ as said) $$\gamma_{k+1}^{-\alpha}\ge \gamma_k^{-\alpha}\big(1- c\\ \gamma_k^\alpha\big)^{-\alpha}\ge \gamma_k^{-\alpha}+\alpha c\\ , $$$$\gamma_{k+1}^{-\alpha}\ge \gamma_k^{-\alpha}\big(1- c\, \gamma_k^\alpha\big)^{-\alpha}\ge \gamma_k^{-\alpha}+\alpha c\, , $$ 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 $C:=(\alpha c ) ^{-1/\alpha}$.
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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 $$\gamma_k=Ck^{-1/\alpha}\big(1+o(1)\big)\\ ,\quad\mathrm{for}\\ k\to\infty \\ .$$$$\gamma_k=Ck^{-1/\alpha}\big(1+o(1)\big)\, ,\quad\mathrm{for}\, k\to\infty \, .$$
Indeed, since $\gamma_k\to 0$, we have $$ \gamma_{k+1}^{-\alpha}= \gamma_k^{-\alpha}\big(1- c\\ \gamma_k^\alpha\big)^{-\alpha}= \gamma_k^{-\alpha}+\alpha c+o(1)\\ ,\quad\mathrm{for}\\ k\to\infty \\ , $$$$ \gamma_{k+1}^{-\alpha}= \gamma_k^{-\alpha}\big(1- c\, \gamma_k^\alpha\big)^{-\alpha}= \gamma_k^{-\alpha}+\alpha c+o(1)\, ,\quad\mathrm{for}\, k\to\infty \, , $$ 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)$$\gamma_{k^{-\alpha}}=k\alpha c+o(k)=k\alpha c\big(1+o(1)\big)$, and the above asymptotic follows.