I have a function $X(t)\geq 0$, with initial condition $X(0)=X_0\geq 0$ and constants $\alpha < 0$, $\beta > 0$ and $\gamma <0$ such that
$$\frac{d}{dt} X(t)^2 \leq \alpha X(t)^2 + \beta X(t) e^{\gamma t}\quad \star$$
If $\beta$ was zero then by Gronwall's inequality, $$X(t)^2 \leq X_0^2e^{\alpha t}$$ so $X(t)$ would go to zero exponentially. Can I say the same for $\star$?
What you have on the left side is $2 X(t) \dfrac d{dt} X(t)$ so that the inequality may be reduced to $$2 \frac d{dt} X(t) \le \alpha X(t) + \beta e^{-\gamma t}.$$ Use the integrating factor $e^{-\alpha t/2}$ to obtain $$\frac d{dt} (X(t)e^{-\alpha t/2} ) \le \frac{\beta}{2} e^{-\alpha t/2} e^{-\gamma t} = \frac \beta 2 e^{-(\alpha /2 + \gamma)t}.$$ If $\alpha + 2\gamma \not= 0$ this leads to $$X(t) \le e^{\alpha t/2} X_0 - \frac{\beta}{\alpha + 2\gamma} \left( e^{-\gamma t} - e^{\alpha t/2} \right)$$ and if $\alpha + 2\gamma = 0$ $$X(t) \le e^{\alpha t/2} \left(X_0 + \dfrac{\beta t}{2} \right).$$