It's rational. However, I am not sure whether or not the denominator is what you think it is.

Let $\Lambda \subset \mathbb{R}^n$ be your lattice.

The covering radius is the smallest $r$ such that every point of $\mathbb{R}^n$ is within $r$ from some lattice point. Let $w$ be a point whose closest distance to $\Lambda$ is exactly $r$.
Let $S$ be the set of points of $\Lambda$ which are $r$ away from $w$.

**Lemma:** $S$ is not contained in any hyperplane.

**Proof:** Suppose, to the contrary, that $S$ is contained in the hyperplane $H$. Let $u$ be a normal vector to $H$. If $w$ is not in $H$, let $u$ point to the side of $H$ on which $w$ lies; if $w$ is in $H$, choose the sign of $u$ arbitrarily. Look at the point $w+\epsilon u$ for small positive $\epsilon$. It is more than $r$ away from every point of $S$. However, if $\epsilon$ is small enough, then it is also more than $r$ away from every point in $\Lambda \setminus S$. So $w+\epsilon u$ is not within $r$ of any point of $\Lambda$, contradicting the definition of $r$. QED

Since $S$ is not contained in a hyperplane, we can choose $n+1$ points of $S$ which do not lie in a hyperplane. Without loss of generality, let one of these points be $0$, and call the others $v_1$, $v_2$, ..., $v_n$. I will show that $r^2$ is rational, and its denominator divides $2^{n+1} \Delta$ where $\Delta$ is the determinant of the lattice generated by the $v_i$'s. However, it is not obvious to me that the lattice generated by the $v_i$ is always $\Lambda$. Moreover, I could imagine that it might happen that the $v_i$'s usually generate $\Lambda$, but every once in a very rare while they don't, which would explain why a numerical search wouldn't find this phenomenon. So I am not sure whether the denominator is exactly what you think it is.

Let's finish the proof. Since the $v_i$ form a basis for $\mathbb{R}^n$, let's write $w = \sum a_i v_i$.

Now, $w$ is equidistant from $0$ and from $v_i$, so $w$ lies on the hyperplane $\{ x: \langle v_i, x \rangle = |v_i|^2/2 \}$. In other words,
$$\sum_j a_j \langle v_i, v_j \rangle = |v_i|^2/2 \quad (*)$$
For every $i$, $(*)$ gives a linear equation in the $a_i$. The right hand side is a half integer. The matrix $\left( \langle v_i, v_j \rangle \right)$ has determinant $\Delta$, and each entry of this matrix is a half integer. So the inverse matrix has entries whose denominators divide $2^{n-1} \Delta$, and we see that the denominators of the $a_i$ divide $2^n \Delta$.

Then
$$r^2 = \langle w,w \rangle = \sum_{i,j} a_{i} a_{j} \langle v_i, v_j \rangle. \quad (**)$$
It is immediately obviously that this is rational.

To get the denominator bound, use $(*)$ to turn $(**)$ into
$$\sum_i a_i |v_i|^2/2.$$
As we saw above, $a_i$ has denominator dividing $2^n \Delta$, and $|v_i|^2/2$ is a half integer. So the denominator of $r^2$ divides $2^{n+1} \Delta$, as I promised.