A very similar, but somewhat easier question, is the following: take $R$ a rectangle, say $[0,m]\times[0,1]$ identifying $\mathbb C$ to $\mathbb R^2$ for nicer notation. This has modulus $m$. Now let $\gamma$ be a simple curve connecting the top to the bottom sides of $R$, while staying within the rectangle. It splits it into two simply connected domains $R_1$ and $R_2$ that are topological rectangles. So each one has a modulus, say the moduli are $m_1$ and $m_2$. Do $m_1$ and $m_2$ determine $m$?
The answer is no here, because of the following two examples.
First, looking at $[0,m_1+m_2]\times[0,1]$ split vertically by $\{m_1\} \times [0,1]$, if the answer was yes, then it would have to be $m=m_1+m_2$.
Second, one can split $[0,m]\times[0,1]$ using a curve that goes back and forth many times between the neighborhood of $\{0\}\times[0,1]$ and the neighborhood ot $\{m\}\times[0,1]$. This will make both $m_1$ and $m_2$ very small, and is incompatible with their sum being equal to $m$.
Something very similar can be done in your case: take $\gamma_1$ straight, and $\gamma_2$ going back and forth between the left and right sides of $\gamma_1$. This will make the moduli of both $A$ and $B$ very small.
One thing which on the other hand is true, in the simply connected case at least, is a sub-additivity relation: with the above notation, one always has $m \geqslant m_1 + m_2$. One way to prove this is via extremal length, as follows. (Not sure the terminology is completely standard.)
Let $\rho : R=[0,m]\times[0,1] \to \mathbb R_+$. The $\rho$-length of a rectifiable curve $\gamma$ is $L_\rho(\gamma) := \int_\gamma \rho ds$ (in terms of the curvilinear coordinate). The $\rho$-width $W_\rho(R)$ of $R$ is the shortest $\rho$-length of a curve connecting the two vertical sides of $R$. The $\rho$-area of $R$ is $A_\rho(R) = \int_R \rho rho^2 |dz|$ (with respect to the Lebesgue measure on $R$). Then, the extremal width of $R$ is defined as $$W(R) := \sup_\rho \frac {W_\rho(R)^2} {A_\rho(R)}.$$
It is easy to define this for any topological rectangle, and to check that $\rho$ is conformally invariant. In particular it has to be a function of the modulus. In the case of the rectangle, the supremum is reached when $\rho$ is constant, in which case $W_\rho(R)=m$ and $A_\rho(R)=m$, leading to $W(R)=m$. In other words: the extremal width is just the same as the modulus.
But now, given a rectangle $R$ split into two rectangles $R_1$ and $R_2$, one can put maximizing functions $\rho_1$ and $\rho_2$ for the previous variational problem on them. This defines $\rho$ on $R$, which does not do better than the optimal one for $R$: the last sentence is exactly the inequality $m \geqslant m_1 + m_2$.
Maybe things have to be tweaked a little bit in your case, because you are gluing objects of different topologies, but the same philosophy will apply. Given more geometrical information, one might also use extremal length to derive better bounds.
Reference for all that: Ahlfors, Lars V. (1973), Conformal invariants: topics in geometric function theory, New York: McGraw-Hill Book Co., MR 0357743.
