I believe this is true.
The argument is that if we fix $p_{XY}$, then the marginal $p_X$ is immediately fixed. So for any $q_X\neq p_X$, there exists no conditional probability distribution $Q_{Y|X}$ such that $q_X Q_{Y|X} = p_{XY}$. This implies that every joint distribution has a unique decomposition into a marginal and conditional distributions.
Is my argument correct?
The answer is no.
Let $(\Omega_X, \mathcal{F}_X)$ and $(\Omega_Y, \mathcal{F}_Y)$ be measurable spaces, and let $X:\Omega_X \rightarrow \mathbb{R}$ and $Y: \Omega_Y \rightarrow \mathbb{R}$ be random variables. Let $P_X$ be a probability measure on $(\Omega_X, \mathcal{F}_X)$. The joint distribution of $X$ and $Y$ is the extension of a pre-measure on the semialgebra $\{A_X \times A_Y \subset \Omega_X \times \Omega_Y: A_X \in \mathcal{F}_X,\ A_Y \in \mathcal{F}_Y \}$ given by $$P((X,Y) \in (A_X \times A_Y)) = \int_{A_X} K(\omega_X, A_Y) P_X(\mathrm{d}\omega_X),$$ where $K:\Omega_X \times \mathcal{F}_Y \rightarrow [0,1]$ is a transition function. In other words, $K$ satisfies
(1) for each $\omega_X \in \Omega_X,$ $K(\omega_X, \cdot)$ is a probability measure on $(\Omega_Y, \mathcal{F}_Y)$; and
(2) for each $A_Y \in \mathcal{F}_Y$, $K(\cdot, A_Y)$ is measurable with respect to $\mathcal{F}_X$, where we assume $[0,1]$ has the Borel $\sigma-$algebra generated from the standard Euclidean topology.
By Caratheodory's Extension theorem, this pre-measure extends uniquely to a probability measure on $\mathcal{F}_X \times \mathcal{F}_Y.$
The "conditional distribution" to which you refer is $K$. It then follows that $K(\cdot, A_Y)$ can take arbitrary values on sets to which $P_X$ assigns measure $0$.
Source: A Probability Path by Sidney Resnick.