Suppose such $\delta<1$ and $\epsilon(\delta)$ exist. Take $A = J + \delta e_1 e_n^T$, $P = e_1 e_1^T$ where $J$ is a nilpotent Jordan block of size $n$, and $e_1,e_n$ are the first and last column of the identity matrix $I$. Since $\|P-Q\|<1$, and orthogonal projection matrices have integer trace equal to their rank, $Q$ must have rank 1. Then $\operatorname{Im} Q$ must be a real eigenvector of $A$, and the only one is $[1\, \delta^{1/n}\, \delta^{2/n}\, \dots \, \delta^{1-1/n}]^T$ (and its multiples). So $Q$ is at a distance $O(\delta^{1/n})$ from $P$. Hence $\epsilon(\delta) > C\delta^{1/n}$ for some $\delta$. This holds for all $n$, and thus $\epsilon(\delta) \geq 1$, contradicting $\epsilon(\delta) \to 0$.

EDIT: I realized that this does not work, because $PA$ has one nonzero too much, sorry. $A$ and $P$ can be simultaneously almost-triangularized, but they don't almost-commute. I'm leaving it up as an attempt, but it doesn't deserve acceptance / upvotes.

Suppose such $\delta<1$ and $\epsilon(\delta)$ exist. Take $A = J + \delta e_1 e_n^T$, $P = e_1 e_1^T$ where $J$ is a nilpotent Jordan block of size $n$, and $e_1,e_n$ are the first and last column of the identity matrix $I$. Since $\|P-Q\|<1$, and orthogonal projection matrices have integer trace equal to their rank, $Q$ must have rank 1. Then $\operatorname{Im} Q$ must be a real eigenvector of $A$, and the only one is $[1\, \delta^{1/n}\, \delta^{2/n}\, \dots \, \delta^{1-1/n}]^T$ (and its multiples). So $Q$ is at a distance $O(\delta^{1/n})$ from $P$. Hence $\epsilon(\delta) > C\delta^{1/n}$ for some $\delta$. This holds for all $n$, and thus $\epsilon(\delta) \geq 1$, contradicting $\epsilon(\delta) \to 0$.