In one direction it's clear: If $P_{X,Y}\ll P_X\otimes P_Y$, then there is a jointly measurable density $f$, and (a version of) the conditional distribution $P_{Y|X=x}$ is given by $f(x,y)P_Y(dy)$, so the choice $\mu=P_Y$ suffices.

Conversely, suppose that $Q$ holds. Then there is a jointly measurable function $g$ such that $P_{Y|X=x}(dy) = g(x,y)\mu(dy)$, for $P_X$-a.e. $x$. Define $h(y)=\int_R g(x,y)\,P_X(dx)$, and observe that if $h(y)=0$ then $g(x,y)=0$ for $P_X$-a.e. $x$. Define $B=\{y: h(y)=0\}$, and $f(x,y) = 1_{B^c}(y)g(x,y)/h(y)$. I claim that $f\cdot P_X\otimes P_Y=P_{X,Y}$. Indeed,
$$
\eqalign{
P_{X,Y}(C)
&=\int\int 1_C(x,y)P_X(dx)P_{Y|X=x}(dy)\cr
&=\int\int 1_C(x,y)P_X(dx)g(x,y)\mu(dy)\cr
&=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr
}
$$
In particular, taking $C=R\times A$, for $A$ a Borel subset of $R$,
$$
\eqalign{
P_Y(A)=P_{X,Y}(R\times A)
&= \int\int 1_A(y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy)\cr
&= \int 1_A(y)1_{B^c}(y)h(y)\mu(dy)\cr
&= \int 1_A(y)h(y)\mu(dy).\cr
}
$$
That is, $P_Y=h\cdot\mu$. Using this to continue the computation in the last-but-one display (multiply and divide by $h(y)$) we have
$$
\eqalign{
P_{X,Y}(C)
&=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr
&=\int\int 1_C(x,y)P_X(dx)f(x,y)P_Y(dy),\cr
}
$$
and the asserted absolute continuity.