Suppose the complex number corresponding to the center of the square is $a$. Then after translating the origin of the complex plane to $a$, the vertices of the square distribute evenly on the circumference. That is, they are the solutions of equation $(x-a)^4 = b$, where $b$ is a complex number. Hence,

$$\begin{align*} x^4 + px^3 + qx^2 + rx + s &= (x-a)^4 - b \\ &= x^4 - 4a x^3 + 6a^2 x^2 - 4a^3 x + a^4 - b. \end{align*}$$

Comparing the coefficients of terms for $x$ with the same degree, we know that $-a = \frac{p}{4}$, and it is a rational number. Combining further that $-4a^3 = r$ is an integer, we can see that $a$ is an integer. So by using the fact that $s = a^4 - b$ is an integer, we can show that $b$ is also an integer.

The above discussion makes clear of a fact that the four numbers corresponding to the four vertices of this square are roots of integer coefficients equation $(x - a)^4 = b$. Hence, the radius of its circumcircle ($=\sqrt[4]{|b|}$) is not less than $1$. Therefore, the area of this square is not less than $(\sqrt{2})^2 = 2$. But the four roots of the equation $x^4 = 1$ are corresponding to the four vertices of a square on the complex plane. Hence, the minimum value of the area of the square is $2$.