This is only a partial answer, as the computations can become quite involved.

We have
$$ \mathbb{E}( X_t J_t ) = \mathbb{E}\left( X_t Q \int_{ \mathbb{R}_+ \times [0, t] } d\mu \right) = \mu_Q \mathbb{E}\left( X_t \int_{ \mathbb{R}_+ \times [0, t] } d\mu \right). $$
by independence of $Q$ with the other processes.

By conditioning on $ \mathcal{F}_t = \sigma(W_1(s), s \leqslant t)\otimes \sigma(W_2(s), s \leqslant t) $, we have
$$ \mathbb{E}( X_t J_t ) = \mu_Q \nu(\mathbb{R}_+) \mathbb{E}\left( X_t \lambda_t \right). $$

The problem is now to understand the product of two CIR processes. An application of Itô formula gives $ d(X \lambda) = X d\lambda + \lambda dX + d\langle X, \lambda \rangle $ and $ \langle X, \lambda \rangle_t = \rho \sigma_1 \sigma_2 \sqrt{X(t) \lambda(t) } dt $. The problem lies in the $ (\theta_1, \theta_2) $. If these parameters are equal to $0$, you get the squared Bessel process which has a semi-group property. Here, you have correlated BM ; expressing the product $ XY = Z $ as a solution of the same SDE seems complicated with $ \theta_i \neq 0 $.

Possible attacks:

1/ You can have the expectation $ f(t) = \mathbb{E}( X_t ) $ (or $ \lambda_t $) by solving the ODE $ f(t) = f(0) + \kappa_1 \int_0^t (\theta_1 - f(s) ) ds $ with $ f(0) = \mathbb{E}(X_0) $ (obtained by taking the expectation of the SDE, with the martingale of expectation 0). You could try to integrate the SDE in $ Z := X\lambda $, of the form $ dZ = (\kappa_1 + \kappa_2) Z dt + \rho \sigma_1 \sigma_2 \sqrt{Z} dt + (\alpha X + \beta \lambda) dt + dM_t $ where $ M $ is the martingale term (and $\alpha, \beta = ...$). But you then need to know the quantity $ g(t) = \mathbb{E}( \sqrt{Z_t} ) $. Maybe another equation using the Itô formula for $ (x, y) \mapsto \sqrt{xy} $. I don't think this is a simple problem though. The literature has maybe some closed form using some properties of the CIR process.

2/ If you can access the semi-group/infinitesimal generator, the theory of diffusions gives you $ \mathbb{E}_{a, b}( X_t \lambda_t) = e^{t \mathcal{L}}f(a, b) $ where $ f(x, y) = xy $ and $ (a, b) $ is the initial value of the process. You can write this generator $ \mathcal{L} $ with the help of the Itô formula (it is of the form $ \frac{\sigma_1^2}{2}x\partial^2_x + \frac{\sigma_2^2}{2}y\partial^2_y + \sigma_1\sigma_2 \rho \partial_{x, y} + ...$), but computing the exponential is complicated. The best way would be to diagonalise it, certainly with bi-variate orthogonal polynomials or special functions (I am not expecting sin and cos to be of any help here) ; since the function $f$ is a simple polynomial, this could work (try a general polynomial to see if the 2-recurrence equation can be solved). If anyone has a reference on that, this would be nice.

The compensated case amounts to do the same computations.