The limit is not zero; it does not exist.
FIRST proof. WLOG let $a=1$. Let $-u(z)$ be your Poisson integral,
(in the numerator of your formula) it is a negative harmonic function in the disc, continuous
in the closed disc, $u(1)=0$, and negative at every other point of the circle.
Let $M(r)=max_{|z|=r}u(z),\; 0\leq r\leq 1$. This is a strictly negative, increasing function on $(0,1)$,
and $M(1)=1$. It is known that $M(r)$ is convex with respect to the logarithm, that is
$$r\frac{dM(r)}{dr}$$ is increasing. This is called Hadamard's Three Circles Theorem.
Thus $M'(1)>0$. Now as $u(re^{i\theta})$ is an even function of $\theta$ decreasing on $(0,\pi)$,
we conclude that $M(r)=u(r)$. This means that there exists a sequence $z_k\to 1$,such that
$z_k<1$, and
$$\frac{u(1)-u(z_k)}{1-z_k}\; \quad\to c>0.$$ So on this sequence your limit
$$\frac{-u(z_k)}{|1-z_k|}\quad =c.$$
On the other hand it is clear that your limit is zero on sequences which tend to $1$ "tangentially$,
that is very close to the circle. So the limit does not exist.
SECOND proof. WLOG $a=1$. Let $v$ be your Poisson integral in the numerator. Let $w$ be the Poisson integral of the same function but replaced by $0$ on the right half of the circle. Then $0 < w < v$ in the open disc. But $w(z)=0$ on an arc of a circle near $1$, so by the Symmetry Principle, $\partial w/\partial r \neq 0$ at the point $1$. As $w$ is positive and $w(1)=0$, this derivative is negative. So $v(r)>w(r)>c(1-r)$ for some $c>0$.
