The answer is *no*, at least in general, as shown by the following counterexample.

Take a double cover $\bar{f} \colon \bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1$, branched over a reducible
curve of the form $B=L_1 + L_2 + M_1 + M_2$ (here $|L|$ and $|M|$ are the two pencil of lines on the quadric).

Such a cover exists because $B$ is $2$-divisible in $\mathrm{Pic}(\mathbb{P}^1 \times \mathbb{P}^1)$, and corresponds to an étale cover $f \colon X \to C_1 \times C_2$, where each $C_i$ is $\mathbb{P}^1$ - {two points}.

If these points are (say) $0$ and $1$ in both factors, then the equation for $X \subset \mathbb{C} \times (\mathbb{C}-\{0, \, 1\})^2$ is

$$z^2 = xy(x-1)(y-1), \quad f(z, (x, \,y)) = (x,\, y).$$

It is clear that the general line in $|L|$ and $|M|$ intersects the branch locus $B$ transversally at two points, hence both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to $\mathbb{P}^1$ (double cover of $\mathbb{P}^1$ branched at two points).

Then both compositions $$X \to C_1 \times C_2 \to C_i$$ have connected fibres, the general one being isomorphic to the $\mathbb{P}^1$ above minus the ramification, i.e. $\mathbb{P}^1$ minus two points, that is clearly connected.

In the same vein, choosing as $B \subset \mathbb{P}^1 \times \mathbb{P}^1$ a divisor of type $$B = \sum_{i=1}^{2g+2} L_i + \sum_{i=1}^{2g+2} M_i,$$
both compositions $$\bar{X} \to \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$$ have connected fibres, the general one being isomorphic to a hyperelliptic curve $\Sigma_g$ of genus $g$, and so both compositions $$X \to C_1 \times C_2 \to C_i$$ (here each $C_i$ is $\mathbb{P}^1$ minus $2g+2$ points) have connected fibres, the general one being isomorphic to $\Sigma_g$ minus $2g+2$ distinct points.