For the (time-homogenous) critical Galton-Watson tree, the survival probability is of order $1/n$. The argument goes through for time-inhomogenous as well, if you're willing to assume that all offspring distributions have expectation $\ge 1$ (and uniform bound on the number of possible offsprings).
A straightforward way of showing this is by induction. Write $q_n$ for the survival probability until time $n$. Then $$q_{n+1}=p_1 q_n + p_2 (2 q_n - q_n^2) \ ,$$ where $p_0,p_1,p_2$ is the offspring distribution at time 0. Let's assume that the expectation is exactly 1 (there's clearly monotonicity here), then $$q_{n+1}=q_n-p_2 q_n \ ,$$ and $p_2\le 1/2$.
Now it a matter of bounding the sequence $q_n$. It will be easier to work with $r_n=1/q_n$, and show that it is growing at most linearly. Then we get $$r_{n+1}=\frac{1}{\frac{1}{r_n}-\frac{p_2}{r_n^2}}=r_n+\frac{p_2}{1-\frac{p_2}{r_n}}$$ so as $r_n\to\infty$ we get that $0\le r_{n+1}-r_n\le p_2 + o(1)$ which is what we want.
EDIT: I see you have edited the question slightly and ask specifically about the case of percolation. In that case, if you assume that all $p_i\ge 1/2$ it is immediate that the survival probability is at least $c/n$ by monotonicity. My answer above is more general and can be easily extended for the case of arbitrary uniform bound on the number of offspring.