The function can be extended on the whole plane as follows:
G(z+na+imb) = F(z) \qquad n,m\in \mathbb Z
to get an analytic bounded function.
added. The resulting function is bounded because it is periodic, hence its range $G(\mathbb C) = G(R) = F(R)$ where $R=[0,b]\times [0,b]$ is the given rectangle, and being $R$ compact and $F$ continuos, $F(G)$ is compact hence bounded. The resulting function is analytic because an analytic function in a point is uniquely determined by the values on a sequence of approximating points. Hence if $F$ is analytic and $F(z)=F(z+a)$ for some point $z$ in the left side of the rectangle, the extension in $F(z+a)$ coincides with the extension inn $F(z)$.
By Liouville Theorem the extended function is bounded and so must be the original function...
note. I assume that the function is analytic on the closed rectangle, which means that there is an open set containing the rectangle where the function has an analytic extension.