In fact, *every* complex manifold has such an atlas.

Let $(M,J)$ be a (finite-dimensional) complex $n$-manifold and let $\mathscr{U}$ be an open cover of $M$ with the properties that *(i)* for each $U\in\mathscr{U}$, there is a $J$-holomorphic chart $\zeta:U\to\mathbb{C}^n$, and *(ii)* For each $U\in\mathscr{U}$ there is a point $p\in U$ that does not lie in any $V\in\mathscr{U}$ other than $U$. (Using paracompactness, it is not difficult to construct such a chart.) Then by choosing one such 'reference point' $p_U\in U$ with $p_U\not\in V\in\mathscr{U}$ for $V\not=U$ and  one $J$-holomorphic chart $\zeta_U:U\to\mathbb{C}^n$ so that $\zeta_U(p_U) = 0\in\mathbb{C}^n$, we arrive at a 'pointed atlas' 
$$
\widehat{\mathscr{U}} = \{ (U,\zeta_U,p_U)\ |\ U\in \mathscr{U}\ \}
$$
with all the stated properties.  The reason is that the only time the point $p_U$ is in the domain of a transition function for the pointed atlas $
\widehat{\mathscr{U}}$ is when one is 'transitioning' from $U$ to $V=U$, and, in that case, the only transition function is the identity mapping on $\zeta_U(U)\subset\mathbb{C}^n$, whose Taylor series at $\zeta_U(p_U) = 0\in\mathbb{C}^n$ clearly has all coefficients in $\mathbb{Q}$ (in fact, all the coefficients are in $\mathbb{Z}$).