So the real question is, given f analytic in the punctured disk $(D_1 \backslash \{ 0\})$ and $\int_{D_1} |f(z)|^{1/4} < \infty$, characterize the smoothness of f at 0.
It is easy to show that if $z=0$ is a pole, then it is at most of order 7.
However, I cannot show that it cannot be an essential singularity. (Or find a counter example)
Remark, if instead we had that $\int_{D_1} |f(z)|^p < \infty$, for $p >=1$, it is relatively easy to show that it cannot be an essential singularity, however it heavily depends on Holder inequality.
Lemma: For an entire non constant function $g$ and $p >0$ we cannot have $\int_{2 \le |z| \le R}|g(z)|^p/|z|^2dA \le A+B \log R$ for positive constants $A,B$ and arbitrary $R >2$
Assume the Lemma and consider $z^8f(z)=\sum_{n=-\infty}^{\infty}a_nz^n=h(z)+k(z)$ the Laurent expansion of $f$ on the unit disc decomposed in the nonpositive power part $h$ and the (strictly) positive power part $k$, so $h(z)=\sum_{n=-\infty}^{0}a_nz^n, k(0)=0$ and $g(w)=h(1/w)$ is entire
Since $k$ is analytic in the unit disc, $\max_{|z| \le 1/2}|k(z)|=C$ finite, so using the inequality $|a|^p+|b|^p \ge |a+b|^p, 0 \le p \le 1$ we get $|h(z)|^{1/4} \le |z|^2|f(z)|^{1/4}+C^{1/4}, |z| \le 1/2$ or
$|h(z)|^{1/4}/|z|^2 \le |f(z)|^{1/4}+C^{1/4}/|z|^2, A_r=r \le |z| \le 1/2$ hence integrating on the annulus $r \le |z| \le 1/2$ we get (using the hypothesis for $f$):
$\int_{A_r}|h(z)|^{1/4}dA_z/|z|^2 \le C_1-C_2\log r$
Changing variables $w=1/z, dA_w=|w'(z)|^2dA_z=dA_z/|z|^4=|w^2|dA_z/|z|^2$ and $B_R$ the annulus $2 \le |w| \le R=1/r$ we get:
$\int_{B_R}|g(w)|^{1/4}dA_w/|w|^2 \le C_1+C_2\log R$, so by the Lemma we get that $g$ is constant, hence $z^8f$ analytic in the unit disc, so the conclusion that $f$ has a pole of order at most $8$, but then we get an easy contradiction with the hypothesis when the order is actually $8$
The Lemma follows from the fact that for any nonnegative number $x$ we have $p\log x \le x^p, p>0$ where $\log 0 = -\infty$ so first assuming $f(0) \ne 0$ we have $p\log |f(w)|-2\log |w| \le |f(w)|^p/|w^2|$ and now integrating on the annulus $B_R$ and using Jensen and the given inequality, it follows that $f$ has finitely many zeroes outside $|z| \ge 2$ hence in the plane and writing $f=Pe^g$ it immediately follows that $g$ must be constant, while for a polynomial $P$ of degree $n \ge 1$ it is clear that $\int_{2 \le |z| \le R}|P(z)|^p/|z|^2dA$ diverges like $R^{np}$ for large $R$ (since $|P|$ behaves like $R^n$ there) so it cannot be $O(\log R)$ and we are done for this case.
But if $f(0)=0$ writing $f(z)=z^kh(z)$ we can repeat the above for $h$ as the inequality only improves after all for large $z$ so we are done in general!