Suppose $A$ is a complete subalgebra of a complete boolean algebra $B$, and $B$ is $\kappa$-centered. Let $G \subset A$ be a generic ultrafilter. Is $B/G$ $\kappa$-centered in $V[G]$?

Naively, we might attempt to prove it as follows. Let $\{ F_\alpha : \alpha < \kappa \}$ be a collection of filters and in $V[G]$ let $F^*_\alpha = \{ [b]_G : b \in F_\alpha, b \not=_G 0 \}$. But for $b_1, b_2 \in F_\alpha$, maybe $b_1 \wedge b_2 =_G 0$, so that $F^*_\alpha$ is not a filter.

An example where the centeredness cardinal goes up when modding out by a filter is given by comparing $\scr P(\omega)$ and $\scr P(\omega)/ \mathrm{fin}$. But maybe the generic filter case is different.