Indeed no such function exists.
a) If $f$ can be extended holomorphically through $0$, the inequality will fail for small $|z|$ since $|z|^{-1/2}$ tends to $\infty$ when $z$ tends to $0$, whereas $f(z)$ tends to $f(0)$.
b) If $f$ can be extended meromorphically through $0$ with a pole of order $k\gt 0$ the function $g(z):=z^kf(z)$ will be entire and satisfy $|g(z) |\geq |z|^{k-1/2} $ so that $h(z)=1/g(z)$ satisfies $|h(z)|\leq |z|^{−k+1/2 }$.
This implies that $h(z)$ can be extended holomorphically at $\infty $ by $h(\infty )=0$ and thus that $g(z)=z^kf(z)=1/h(z)$ has a pole ay $\infty$.
Hence $f(z)$ is meromorphic on the whole extended plane and is thus a rational function , necessarily of the form $f(z)= \frac {1}{z^k} P(z)$ with $P$ a polynomial satisfying $P(0)\neq 0 $.
But then:
$\bullet$ if $P$ is not constant our inequality is false at any zero of $P$.
$\bullet$ if $P$ is constant our inequality is false for large $|z|$.
c) If $f(z)$ has an essential singularity at $0$, then in the pointed disk $D^* $ defined by $0\lt |z|\lt 1$ the inequality implies $|f(z)| \gt 1$ .
As you very correctly conjectured, this contradicts the big Picard theorem according to which $f$ takes all complex values (except maybe one) in that pointed disk,.