Certainly not always. The most trivial example seems to be $X=\ell^\infty$, $\eta_n=\pm e_n$ (with probability $1/2$ for each sign), and $\xi_n$ being uniformly distributed on $\pm e_1,\dots,\pm e_N$ with large $N$ for $n=1,\dots,N$ (as usual, $e_n$ is the vector with the $n$-th coordinate $1$). The rest of $\xi_n$ and $\eta_n$ can be put to $0$, say.

Then the sums of probabilities in question are equal (and equal to one half times the number of vectors $\pm e_n, n=1,\dots,N$ lying in $A$). However, $\|\sum\eta_n\|$ is always $1$ while $\|\sum\xi_n\|$ is typically (i.e., with probability $\ge \frac12$, say) at least of order $\sqrt{\frac{\log N}{\log\log N}}$ (that is just the classical balls into bins problem combined with the random choice of signs for the balls in the maximal bin, and in this computation I ignore the fact that the bin with nearly maximal number of balls is typically not unique, which may drive the estimate up even more).

Next, the "perhaps, depending on $X$" construct looks very fishy in the question as it is posed now (nobody prevents us from taking the appropriate sum of spaces with large constants to get a space without any constant, say; also, since the question is essentially finite-dimensional, it is clear that $\ell^\infty$ must be a counterexample if anything at all is).

Are you sure that the problem is not "Describe (in some familiar terms) the Banach spaces $X$ such that..."?