I guess that the answer is no in general. More precisely what I consider as the discrete version of your question has a negative answer. I guess that one should be able to find a couterexample to your question by an ultraproduct argument, but I did not check the details.

By discrete version of your question I mean: if $(P_n)_{n \in \mathbb N}$ is a family of uniformly bounded projections on $X$ such that $P_0=0$ and $P_n P_m = P_{\min(n,m)}$, and $Q$ is a (not necessaily bounded) linear map on $X$ that commutes with the $P_n$'s and is such that $\sup_n \|P_n Q -P_{n-1} Q\|<\infty$. Does it follow that $\sup_n \|P_n Q\|<\infty$?

If $(e_n)$ is a Schauder basis in $X$, and take $P_n$ the map $x = \sum_k x_k e_k \mapsto P_n(x) = \sum_{k \leq n} x_k e_k$. Then $P_n$ satisfies your assumption. For a sequence $z_n \in \{{-1,1\}}$, define $Q x_n = z_n x_n$ so that $\|P_n Q - P_{n-1} Q\| = \|P_n - P_{n-1}\|$. Then $\sup_n \|P_n Q\| <\infty$ for every such $Q$ is equivalent to the basis $(e_n)$ being unconditional. It is well known that there exist bases that are not unconditional. This answers negatively the discrete question.

Here is how I would try to deduce a continuous example from a discrete example $X,P_n,Q$~: for any $k$, define $P_s^{(k)} = P_{[2^k s]}$ and $Q_k = P_{2^k}Q/\|Q P_{2^k}\|$. Then consider a non principal ultrafilter on $\mathbb N$, and construct the following operators on $X^{\mathcal U}$, and the operators $P_s = (P_s^{(k)})_{k \in \mathbb N}$ and $Q_{\mathcal U} = (Q_k)_{k \in \mathbb N}$. Then $\|Q_{\mathcal U}\|=1$, so that $P_0 Q_{\mathcal U} = 0 \neq Q_{\mathcal U}= P_1 Q_{\mathcal U}$. One probably needs to assume something more on $Q$ to prove that $s \mapsto P_s Q$ is continuous. I leave this to you.