Tue Tjur studied the existence of continuous disintegrations in a 1975 preprint "A Constructive Definition of Conditional Distributions," Issue 13, Copenhagen Universitet. He gives necessary and sufficient conditions for their existence, at least in the setting of Radon measures.
He also discusses sufficient structure, and there considers a basic probability space that is an open subset of a finite-dimensional Euclidean space and the problem of conditioning on a random variable taking values in an open subsets of a Euclidean space $\mathbb R^k$ such that, when the random variable is considered as a (measurable) function, it is surjective and continuously differentiable with differential of maximal rank. This particular special case may be too narrow for you, but perhaps the general case can give you some guidance.
The article is a bit hard to track down, so let me know if you need help finding it. The existence of continuous disintegrations arises also in the study of the computability of conditional probability, which is my interest.
Let's assume that we are working with the canonical probability space $\Omega = D(\mathbb R)$ of càdlàg functions, and $\mathbb P$ is the law of the process. I would doubt that there is a satisfactory answer at the level of maximal generality you've stated. At the very least, the measure $\mathbb P$ should be Radon. There are extremely general results on the existences of RCPs for Radon measures (cf. Leão, Fragoso and Ruffino, Regular conditional probability, disintegration of probability and Radon spaces).
The RCP is a measure-valued function $P : [0,T] \times \Omega \to \mathcal M(\Omega)$ such that for $\mathbb P$-almost every $\omega$, the measure $P(t,\omega, \cdot)$ is a version of $\mathbb P(\cdot|\mathcal F_t)$. Do you want the function $(t, \omega) \mapsto P(t,\omega,\cdot)$ to simply exist and be measurable? If so, this can be done in the wide generality stated above; see Leão et al.
Recently, I have needed more regularity properties for RCPs, namely, continuity. Consider the space $\mathcal M(\Omega)$ of Radon measures on $\Omega$ equipped with the topology of weak convergence of measures. We say that the RCP is a continuous disintegration (or continuous RCP) when it satisfies the following property: $$\mbox{if $\omega_n \to \omega$, then the measures $P(t,\omega_n,\cdot)$ converge weakly to $P(t,\omega,\cdot)$.}$$
If the law is Gaussian, then my preprint Continuous Disintegrations of Gaussian Processes gives a necessary and sufficient condition for the law $\mathbb P$ to have a continuous disintegration. I haven't thought about this in the case of càdlàg functions, but I'm pretty sure that this will extend easily. Note that this is just for fixed $t$.
To show that the map $(t, \omega) \mapsto P(t,\omega,\cdot)$ jointly continuous, a little more work is needed. As part of a larger project, Janek Wehr and I have general results in this direction for stationary, Gaussian processes. If this is what you need, I'm happy to discuss this with you further.
Open Question: If the law $\mathbb P$ is not Gaussian but at least is log-Sobolev, then all the same results should hold. This is because log-Sobolev measures satisfy very strong concentration-of-measure properties. I have some ideas how to do this, but I haven't worked out the details because I've been busy with other projects. If anybody is interested in collaborating on extending this work to the log-Sobolev case, please contact me.
Best Answer
Even if your probability measure is absolutely continuous with respect to Lebesgue measure on $\Omega={\mathbb R}^2$, I don't think this suffices for the function $f$ you have defined to be continuous (just take $X$ to be independent from $Y$, i.e. your probability measure is just the product of two probability measures, "one on each axis", and choose the one for $X$ to be something in $L^1({\mathbb R}, {\mathcal B}, dx)$ which is discontinuous.
On the other hand, if ${\mathbb P}$ is not just absolutely continuous with respect to Lebesgue measure on $\Omega$, but has a continuous density function wrt said measure, then your function $f$ will be continuous -- just because integrating over a ball of radius $\eta$ with centre $\eta$ can only smooth things out, so that continuity of the original density function goes over to continuity of your conditional probability. [This is a fairly straightforward observation using basic properties of usual integration in the plane.]
So in your example, the Gaussian structure isn't really relevant as far as I can see. Also, if the original density function is strictly positive on the cylinder $\{(x,y) : |y-y_0|<\eta \}$, then the conditional probability you've defined will also be strictly positive; this condition is evidently not necessary, but I suspect in the examples you're interested in something like it should hold.
In between these two extremes, I'm not sure what else one can say. Perhaps, from your point of view, it's more important to go up to infinite-dimensional $\Omega$ but place restrictions on the kind of probability measure which you wish to consider.