On sums of independent random variables in Banach spaces
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..."?
First of all, many thanks to @LvivScottishBook for putting the problem online (that was indeed very surprising for me) and to @fedja for finding a counterexample.
While writing the problem I was sure that it has a positive answer for any Banach space (for some reason this was kind of intuitive for me) with $C$ independent of $X$, so @fedja answered the original question. Nonetheless, as @fedja noticed, it is not clear for which Banach spaces such an inequality holds true. The only thing that I can guarantee so far is that the necessary assumption on the Banach space is having a finite cotype thanks to @fedja counterexample and the Maurey–Pisier theorem (see Corollary 7.3.14 in Analysis in Banach spaces, Volume II by Hytönen, van Neerven, Veraar, and Weis).
It remains open whether finite cotype is a sufficient condition as well.