Note that this is a correct answer to the original question, so I will leave it here, even though the question has now been changed. (The original question is recoverable by going back to the previous versions.)
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 $\mathbb{Z}$).