Consider $g(z) = 1/(1-z)^2 = \sum_{n=0}^\infty (n+1) z^n$, which satisfies $\int_D |g(z)| \ dx\ dy = \infty$. Let $S_{N,M}(z) = \sum_{N \le n < M} (n+1) z^n$ and $B_r(q) = \{z \in D: |z - q| < r\}$. For any $r > 0$ and any $N$ and $R$, since $\int_{B_r(1)} |S_{N,\infty}(z)|\ dx\ dy = \infty$, for $\epsilon > 0$ sufficiently small we have
$\int_{B_r(1) \cap B_{1-\epsilon}(0)} ||S(N,\infty)(z)|\ dx\ dy > 2 R$, and we can take $M$ large enough that $\int_{B_r(1) \cap B_{1-\epsilon}(0)} |S(N,M)(z)|\ dx\ dy > R$.

Enumerate pairs $(r, q)$ where $1/r$ is a positive integer and $q = \exp(i t)$ where $t$ is rational as $(r_k, q_k)$. Now define $f(z) = \sum_{k=1}^\infty S_{N_k, N_{k+1}}(z/q_k)$ where $N_k$ is an increasing sequence and $\epsilon_k>0$ a decreasing sequence, defined inductively so that
$$\int_{B_{r_k}(1) \cap B_{1-\epsilon_k}(0)} |S_{N_k,N_{k+1}}(z)| \, dx \, dy> k +
\sum_{j < k} \int_D |S_{N_j,N_{j+1}}(z)|\, dx \, dy$$
and $\sum_{n \ge N_{k+1}} (n+1) (1-\epsilon_k)^n < 1/{2 \pi}$. The sum converges, uniformly on compact sets, to an analytic function on $D$ with

\begin{eqnarray*}
&\int_{B_{r_k}(q_k) \cap B_{1 - \epsilon_k}(0)} |f(z)|\, dx \, dy \cr
&\ge \int_{B_{r_k}(q_k) \cap B_{1 - \epsilon_k}(0)} (|S_{N_k,N_{k+1}}(z/q_k)| -
\sum_{j < k} |S_{N_j,N_{j+1}}(z/q_k)| - \sum_{j \ge N_{k+1}} (n+1) (1-\epsilon_k)^j)\, dx\, dy\cr
&\ge k - 1\cr
\end{eqnarray*}

Since every neighbourhood $U$ of any $q \in S^1$ contains infinitely many $B_{r_k}(q_k)$, $f$ is not in $L^1(U)$.