Answer: $\frac{m^2}{2}$.
Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. It is easy to see that the perpendicularity of the lines $AC$ and $BD$ is equivalent to the equality $(a+c)(b+d) = -1$ and the condition about the difference of projections is equivalent to the equality $(a+c)-(b+d)=m$. The fact that the point $F$ belongs to the lines $AC$ and $BD$ means that $ac = bd = -\frac{1}{4}$. Let $a+c = p$ and $b+d = q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2-px-\frac{1}{4}=0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2-qx-\frac{1}{4}=0$. Hence $c-a = \sqrt{p^2+1}$ and $d-b = \sqrt{q^2+1}$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the lengths of the diagonals. So,
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$$\begin{aligned} S &= \frac{1}{2}(c-a)(d-b)\sqrt{1+(c+a)^2}\sqrt{1+(d+b)^2} = \\ &= \frac{1}{2}(1+p^2)(1+q^2) = \frac{(q+qp^2)(p+pq^2)}{2pq} = \frac{-(p-q)^2}{-2} = \frac{m^2}{2}. \end{aligned}$$