If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$, and $v \in H$ is a cyclical and separating vector for $\mathcal{R}$ (hence also for its commutant $\mathcal{R}'$), and $P \in \mathcal{R}, Q \in \mathcal{R}'$ are nonzero projections, can we have $PQv = 0$?
[note i had briefly edited this to a reformulated version of the question, but am rolling it back to align with the answer below.]
On
Following the suggestion in a comment above, I'll attempt to show that the answer holds for factors too.
Claim: Whenever $\mathcal{R}$ is a factor, acting on a Hilbert space $H$ with dimension $> 1$, having a separating and cyclical vector $v \in H$, there exists a sep./cyc. vector for $\mathcal{R}$ that is annihilated by some $PQ$ as in our question.
Fix $v$ and arbitrary projections $P \in \mathcal{R}, Q \in \mathcal{R}'$ that are nontrivial (i.e. $\neq 0$ and $\neq 1$). (Such $P, Q$ must exist since otherwise either $\mathcal{R}$ or $\mathcal{R}'$ would equal $B(H)$, which has no separating vectors unless $H$ is one-dimensional, a case we have excluded.)
Let $v_1 = PQv$, and $v_2 = P^\perp Q^\perp v$. I believe $P \mathcal{R} P Q$ and $Q \mathcal{R}' Q P$ are both von Neumann algebras acting on $PQH$. If $v_1$ fails to be separating for either algebra, then some nonzero $P' Q$ with $P' \in \mathcal{R}, P' < P$, or else some nonzero $P Q'$ with $Q' \in \mathcal{R}', Q' < Q$, would annihilate $v$, immediately giving us witnesses for our claim.
So suppose $v_1$ is separating for both, and similarly suppose $v_2$ is separating for both $P^\perp \mathcal{R} P^\perp Q^\perp$ and $Q^\perp \mathcal{R}' Q^\perp P^\perp$.
Proposition: $v_1 + v_2$ is separating for both $\mathcal{R}$ and $\mathcal{R}'$. If this is true, then we have witnesses for our claim, since $P Q^\perp (v_1+v_2) = 0$.
Suppose towards a contradiction that $v_1 + v_2$ is not separating for $\mathcal{R}$ (the argument for $\mathcal{R}'$ will be symmetrical). Then some nonzero $X \in \mathcal{R}$ annihilates $v_1 + v_2$.
Now since $X (v_1 + v_2) = 0$ we have $X v_1 = - X v_2$, but the range of a projection is stable under the action of the commutant, and $v_1 \in range(Q)$ and $v_2 \in range(Q^\perp)$. So $X v_1 = - X v_2$ would be impossible unless both $X v_1$ and $X v_2$ were zero. So they are both zero.
Claim: $X v_1 = 0$ entails that $range(P) \subseteq ker(X)$. Suppose otherwise. Then $XP$ is a nonzero operator, and the projection $Z$ onto $(ker(XP))^\perp$ exists in $P \mathcal{R} P$ (by polar decomposition in $\mathcal{R}$) and is nonzero. Now $P \mathcal{R} P$ is *-isomorphic to $P \mathcal{R} P Q$ via the map $T \mapsto T Q$ (Kadison/Ringrose, prop. 5.5.5). So $Z Q$ is a nonzero projection in $P \mathcal{R} P Q$. But we have supposed that $v_1$ is separating for that algebra. So $Z Q v_1 \neq 0$, and since $Q v_1 = v_1$, also $Z v_1 \neq 0$, so $v_1$ is not in the kernel of $X$, contradiction.
Thus $range(P) \subseteq ker(X)$, and a symmetrical argument starting from $X v_2 = 0$ shows that $range(P^\perp) \subseteq ker(X)$. These two facts entail $X = 0$, contradicting our choice of it as a nonzero operator. Thus our supposed $X$ cannot exist, and $v_1 + v_2$ is separating for $\mathcal{R}$. Again, a symmetrical argument shows it is separating for $\mathcal{R}'$.
Answering the original version of the question. Let $H=\mathbb C^2$, and $$ \mathcal R=\left\{\begin{bmatrix} a&0\\0&b\end{bmatrix}:\ a,b\in\mathbb C\right\}\subset M_2(\mathbb C),\ \ \ \ \ \ v=\begin{bmatrix} 1\\1\end{bmatrix}. $$ Then $v$ is cyclic and separating, and $\mathcal R'=\mathcal R$. Take $$ P=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\ \ \ Q=\begin{bmatrix} 0&0\\0&1\end{bmatrix}. $$ Then $P\in\mathcal R$, $Q\in\mathcal R'$, and $PQ=0$, so $PQv=0$.