I don't have a complete answer, but the regularity of this map depends on the Hessian of $H$ at $p$.
I will see $u$ as a solution on the cylinder $\mathbb{R} \times S^1$, where $s$ is the coordinate on $\mathbb{R}$ and $t \in \mathbb{R}/\mathbb{Z}$.
You want to extend $u$ across $-\infty$ by seeing $u(s,t) = U(\operatorname{e}^{2\pi(s+it)})$ for a function $U$ on $\mathbb{C}$.
The solution $u$ has an asymptotic series expansion in terms of eigenfunctions of the asymptotic operator. In the special case that $M = \mathbb{C}^n$, $J = i$, $p = 0 \in \mathbb{C}^n$, the asymptotic operator is given by
$A := -( i \frac{d}{dt} + D^2H(0) )$, acting on maps from $S^1 \to \mathbb{C}^n$.
(Where $D^2H(0)$ is the Hessian of $H$ at $0$.) See this as an unbounded self-adjoint operator on $L^2(S^1, \mathbb{C}^n)$. If you have an eigenvalue $\lambda$ for this operator and a corresponding eigenvector $v(t)$, then the leading order term in the asymptotic expansion for $u$ has a term $\operatorname{e}^{\lambda s} v(t)$. Now, if $U$ were $C^1$, say, it would have a Taylor expansion, notably meaning that the first order term has an integer power. In particular then, if this $0 > \lambda > -2\pi$, your map $U$ won't be $C^1$, but will instead be Hölder continuous for some exponent related to this $\lambda$. I haven't done the computations carefully, but it seems to me that if you assume that $H$ is locally $H(z_1, \dots, z_n) = c( |z_1|^2 + \dots + |z_n|^2)$, the condition on $c$ that $0$ be a non-degenerate periodic orbit precisely rules out the cases when the eigenvalues are in $2\pi \mathbb{Z}$. I suspect this is a more general phenomenon, but I don't see why at the moment.
This type of expansion is discussed in several places. I like the exposition in Robbin-Salamon, Asymptotic behaviour of holomorphic strips. The improvement of the leading order term to higher order terms is done in Siefring Relative asymptotics of pseudoholomorphic half-cylinders.