This seems to be a standard exercise. Anyway, here is a sketch of the solution.
Let $T$ be the hitting time of zero, and $\phi_n(s) = \mathbb{E}(e^{-s T} | X_0 = n)$ be the Laplace transform of $T$. Then $\phi_0(s) = 1$, and $$ \phi_n(s) = \frac{\mu + \theta}{\mu + \theta + s} \left( \frac{\mu}{\mu + \theta} \, \phi_{n+1}(s) + \frac{\theta}{\mu + \theta} \, \phi_{n-1}(s) \right) = \frac{\mu \phi_{n+1}(s) + \theta \phi_{n-1}(s)}{\mu + \theta + s} \, .$$ Solving this system of linear equations (given $0 \le \phi_n(s) \le 1$) leads to $$ \phi_n(s) = \left( \frac{2 \theta}{\mu + \theta + s + \sqrt{(\mu + \theta + s)^2 - 4 \mu \theta}}\right)^n . $$ The Laplace transform of $T = T_A - T_B$, with independent $T_AA$ and $T_B$, is $\phi_A(s) \phi_B(-s)$. The probability that $T > 0$ can be expressed as $$ \frac{1}{2} - \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{\phi_A(i s) \phi_B(-i s)}{s} ds ,$$ with the integral understood in the principal value sense.