Seems that you need to look at how you construct your conditional probabilities.
The answer is yes, provided you use a construction of the conditional probability with "nets", as defined in Feller, An introduction to probability theory and its applications, Vol. 2 (1971), Section V.10. I will require that both $\mathcal X$ and $\mathcal Y$ have a net (which is true if $\mathcal X$ and $\mathcal Y$ are Polish spaces).
Preliminary: "nets", Radon–Nikodym derivatives, and conditional probability kernels
For a space $\mathcal U$, a "net" (in the sense of footnote 20) is a collection $(U_{i_1, \dots, i_n})_{(i_j)\in \{0,1\}^n, n\geq 0}$ such that
- $U_\emptyset = \mathcal U$,
- $(U_{i_1, \dots, i_{n-1}, 0}, U_{i_1, \dots, i_{n-1}, 1})$ is a partition of $U_{i_1, \dots, i_{n-1}}$,
- $\cap_{n\geq 1} U_{i_1, \dots, i_n}$ is a singleton for every sequence $(i_j)_{j\geq 1}$.
When $\mathcal U$ is equipped with a sigma-algebra $\mathfrak U$, I assume in addition that $\mathfrak U$ is generated by the net.
Given a set $\mathcal U$ with a net $(U_{i_1, \dots, i_n})$ generating the sigma-algebra $\mathfrak U$ and a measurable space $(\mathcal V, \mathfrak V)$ with a distinguished (arbitrary) element $v$, for every measurable set $V\in\mathfrak V$ and every probability measure $\mu$ on $(\mathcal U \times \mathcal V, \mathfrak U \otimes \mathfrak V)$, the Radon-Nikodym density of $\mu(\cdot \times V)$ with respect to $\mu(\cdot \times \mathcal V)$ is the $\mu(\cdot\times\mathcal V)$-a.e. limit (with values in $[0,1]$) of the functions
$$ f^V_n : u \in U_{i_1, \dots, i_n} \mapsto
\begin{cases}
\frac{\mu(U_{i_1, \dots, i_n} \times V)}{\mu(U_{i_1, \dots, i_n} \times \mathcal V)} &\text{ if } \mu(U_{i_1, \dots, i_n} \times \mathcal V)>0 , \\
[v\in V] &\text{ otherwise,}
\end{cases} $$
where $[v\in V]=1$ if $v\in V$ and $0$ otherwise. (We can see this by an elegant martingale argument.) Being simple functions, the $f^V_n$ are measurable, hence their a.s. limit $g^V$ is also measurable.
If $\mathcal V$ has a net $(V_{i_1, \dots, i_n})$ generating $\mathfrak V$, then (following Footnote 20) we can construct a Markov kernel
$$\kappa : \mathcal U \times \mathfrak V \to [0,1] , (u,V) \mapsto \kappa(u, V) $$
such that $\kappa(u, \cdot)$ is a probability measure on $(\mathcal V, \mathfrak V)$ for every $u\in \mathcal U$ and $\kappa(\cdot, V)$ is measurable for every $V\in \mathfrak V$: to do so, first fix $\kappa(u, V_{i_1, \dots, i_n, 0}) = g^{V_{i_1, \dots, i_n, 0}}(u)$ and $\kappa(u, V_{i_1, \dots, i_n, 1}) = g^{V_{i_1, \dots, i_n}}(u) - g^{V_{i_1, \dots, i_n, 0}}(u)$, then define for every $V\in\mathfrak V$
$$ \kappa(u,V) = \inf_{n} \sum_{(i_1, \dots, i_n)\in \{0,1\}^n} \kappa(u, V_{i_1, \dots, i_n}) [V_{i_1, \dots, i_n} \cap V \neq \emptyset] $$
and note how $\mu(\cdot\times\mathcal V)$-a.e. we have $\kappa(u,V) = g^V(u)$ (the expression with the infimum allows us to have the measurability of $\kappa(\cdot, V)$ jointly for every measurable $V$). This can be rewritten
$$ \kappa(u,V) = \inf_n \lim_k \sum_{(i_1, \dots, i_n)\in \{0,1\}^n} \tilde f_k^{V_{i_1, \dots, i_n}}(u) [V_{i_1, \dots, i_n} \cap V \neq \emptyset] $$
where $\tilde f_k^{V_{i_1, \dots, i_{n-1}, 0}} = f_k^{V_{i_1, \dots, i_{n-1}, 0}}$ and $\tilde f_k^{V_{i_1, \dots, i_{n-1}, 1}} = f_k^{V_{i_1, \dots, i_{n-1}}} - f_k^{V_{i_1, \dots, i_{n-1}, 0}}$.
Answer.
Let $(\mathcal X, \mathfrak X)$ and $(\mathcal Y, \mathfrak Y)$ be measurable spaces, and let $\mathfrak R$ be the Borel sigma-algebra of $\mathbb R$. We assume that $\mathcal X$ and $\mathcal Y$ come with a distinguished point $x$ and $y$ respectively. I assume that $\mathcal X$ has a net $(A_{i_1, \dots, i_n})$ and $\mathcal Y$ has a net $(B_{i_1, \dots, i_n})$, that generate $\mathfrak X$ and $\mathfrak Y$ respectively.
We clearly have such a net for $\mathbb R$, see footnote 20, and I will denote the elements of this net by $C_{i_1, \dots, i_n}$.
Clearly, the sets $D_{i_1, \dots, i_{2n}} = A_{i_1, i_2, \dots, i_{n}} \times C_{i_1, i_2, \dots, i_{n}}$ and $D_{i_1, \dots, i_{2n+1}} = A_{i_1, i_2, \dots, i_{n+1}} \times C_{i_1, i_2, \dots, i_{n}}$ form a net of $\mathcal X \times \mathbb R$; a net of $\mathcal X \times \mathcal Y$ can similarly be constructed.
From this, we can construct the following Markov kernels:
$\kappa(y, A) = \mathbb P((X,Z)\in A \ | \ Y=y)$ with $y\in Y$ and $A \in \mathfrak X \otimes \mathfrak R$,
for every $y\in Y$, starting from the probability measure $\kappa(y, \cdot)$, the kernel $\kappa_y(x,A) = \kappa_y(Z \in A \ | \ X=x)$ for every $x\in X$ and $A\in\mathfrak R$.
Your question is: Is $(x,y) \mapsto \kappa_y(x, C)$ measurable for every measurable subset $C$ of $\mathbb R$? (This is equivalent to your formulation with $Z\in L^1$ by classical arguments of approximation of $L^1$ random variables by simple random variables.)
By construction,
$$ \kappa_y(x,C) = \inf_n \lim_k \sum_{(i_1, \dots, i_n)\in \{0,1\}^n} \tilde f_k^{C_{i_1, \dots, i_n}}(x,y) [C_{i_1, \dots, i_n} \cap C \neq \emptyset] , \qquad (1) $$
where for every $C \in \mathfrak R$
$$ f^C_k(\cdot, y) : x \in A_{j_1, \dots, j_k} \mapsto
\begin{cases}
\frac{\kappa(y, A_{j_1, \dots, j_k} \times C)}{\kappa(y, A_{j_1, \dots, j_k} \times \mathbb R)} &\text{ if } \kappa(y, A_{j_1, \dots, j_k} \times \mathbb R)>0 , \\
[0\in C] &\text{ otherwise}
\end{cases} $$
and the $\tilde f_k^{C_{i_1, \dots, i_n}}(\cdot, y)$ are defined from the $f_k^{C_{i_1, \dots, i_n}}(\cdot, y)$ as before, so that they are $\mathfrak X \otimes \mathfrak Y$-measurable by the measurability of $\kappa(\cdot, D)$ for every $D\in\mathfrak X \otimes \mathfrak R$. From $(1)$ we conclude that $(x,y) \mapsto \kappa_y(x, C)$ is $\mathfrak X \otimes \mathfrak Y$-measurable for every $C\in\mathfrak R$, finishing the proof.
