No. Weight enumerators with the given normalization are not necessarily integers for non-stabilizer codes. In fact, the first discovered non-stabilizer code provides a counterexample. However, the logical subspace of that code is six-dimensional. See this paper for a $(\!(11,2)\!)$ code which is a counterexample encoding a logical qubit (thank you to
Ian Gershon Teixeira for pointing out this paper).
$(\!(5,6)\!)$ counterexample
The first discovered non-stabilizer code is defined by the following projector
$$
\begin{align}
\Pi=\frac{1}{16}[
\,3\,&I\otimes I\otimes I\otimes I\otimes I\\
+\,&(I\otimes Z\otimes Y\otimes Y\otimes Z)_{\text{cyc}}\\
+\,&(I\otimes X\otimes Z\otimes Z\otimes X)_{\text{cyc}}\\
-\,&(I\otimes Y\otimes X\otimes X\otimes Y)_{\text{cyc}}\\
+2\,&(Z\otimes X\otimes Y\otimes Y\otimes X)_{\text{cyc}}\\
-2\,&Z\otimes Z\otimes Z\otimes Z\otimes Z]
\end{align}\tag1
$$
where the subscript "cyc" indicates the presence of all five cyclic shifts. We have
$$
\begin{align}
A_4&=\frac{1}{(2^{\log_2 6})^2}\sum_{p\in P_5,\mathrm{wt}(p)=4}\mathrm{tr}(p\Pi)^2\tag2\\
&=\frac{1}{6^2}\frac{1}{16^2}(5+5+5)\,\mathrm{tr}(I\otimes I\otimes I\otimes I\otimes I)^2\tag3\\
&=\frac{15}{6^2}\frac{32^2}{16^2}=\frac{5}{3}.\tag4
\end{align}
$$
Admittedly, this code does not match the parameters of the question exactly since its logical subspace is six-dimensional.
$[\![4,2]\!]$ is not a counterexample
Let $C$ denote the joint $+1$ eigenspace of $Y^{\otimes 4}$ and $H^{\otimes 4}$. Any single-qubit Pauli error anti-commutes with one of the two operators, so $C$ is a code that can detect all single-qubit Pauli errors. The projector is
$$
\begin{align}
\Pi&=\frac{I^{\otimes 4}+Y^{\otimes 4}}{2}\frac{I^{\otimes 4}+H^{\otimes 4}}{2}\tag5\\
&=\frac14[I^{\otimes 4}+Y^{\otimes 4}+H^{\otimes 4}+Y^{\otimes 4}H^{\otimes 4}]\tag6\\
&=\frac14\left[I^{\otimes 4}+Y^{\otimes 4}+\frac14(X+Z)^{\otimes 4}+\frac14Y^{\otimes 4}(X+Z)^{\otimes 4}\right]\tag7
\end{align}
$$
where $(X+Z)^{\otimes 4}$ expands to the sum of all sixteen weight four Pauli strings consisting solely of $X$ and $Z$. Let us denote this set with $Q_n=\{P\in P_n\,|\,\mathrm{wt}(p)=n\wedge\mathrm{wt}_Y(p)=0\}$ where $\mathrm{wt}_Y(p)$ denotes the number of qubits on which $p$ acts as $Y$. Now, $YX=-iZ$ and $YZ=iX$,so
$$
\begin{align}
\Pi&=\frac14\left[I^{\otimes 4}+Y^{\otimes 4}+\frac14(X+Z)^{\otimes 4}+\frac14Y^{\otimes 4}(X+Z)^{\otimes 4}\right]\tag7\\
&=\frac{1}{16}\left[4I^{\otimes 4}+4Y^{\otimes 4}+\sum_{q\in Q_4} q+Y^{\otimes 4}\sum_{q\in Q_4} q\right]\tag8\\
&=\frac{1}{16}\left[4I^{\otimes 4}+4Y^{\otimes 4}+\sum_{q\in Q_4} q+\sum_{q\in Q_4} (-1)^{\mathrm{wt}_Z(q)}q\right]\tag9\\
&=\frac{1}{16}\left[4I^{\otimes 4}+4Y^{\otimes 4}+2\sum_{r\in R_4} r\right]\tag{10}\\
\end{align}
$$
where
$$
\begin{align}
R_4=\{
&XXXX, ZZZZ,\\
&XXZZ, ZZXX,\\
&XZXZ, ZXZX,\\
&ZXXZ, XZZX\}\tag{11}
\end{align}
$$
is the set of all eight four-qubit Pauli strings made up of an even number of $X$, an even number of $Z$ and no identity or $Y$. Having found the expansion of the code subspace projector in the Pauli basis, we are ready to compute weight enumerators
$$
\begin{align}
A_4&=\frac{1}{(2^2)^2}\sum_{p\in P_4,\mathrm{wt}(p)=4}\mathrm{tr}(p\Pi)^2\tag{12}\\
&=\frac{1}{4^2}\frac{1}{16^2}(4^2+8\cdot 2^2)\,\mathrm{tr}(I\otimes I\otimes I\otimes I)^2\tag{13}\\
&=\frac{48}{4^2}\frac{16^2}{16^2}=3\tag{14}
\end{align}
$$
where we used $k=2$ which follows from $\mathrm{tr}(\Pi)=4$.
