The question linked by @AlonAmit in the comments answers exactly this question, and shows that the answer (at least with the constant $1/4$) is no. For a concrete demonstration, start with a $6\times 6$ square broken into thirty-sixths:
Now cover each corner $2\times2$ by four individual $(1+\varepsilon)\times (1+\varepsilon)$ rectangles, such that the four rectangles in the upper left ($0,1,6,7$) are mutually overlapping, as are those in each of the other corners. And cover the remaining shape in the center (a cross) by eight individual $(3+\varepsilon)\times(1+\varepsilon)$ rectangles ($28e$, $39f$, $cde$, $ijk$, $kqw$, $lrx$, $fgh$, and $lmn$), such that all eight include the center of the square. Any disjoint set of these rectangles includes at most four of the $(1+\varepsilon)\times(1+\varepsilon)$ rectangles and at most one of the $(3+\varepsilon)\times(1+\varepsilon)$ rectangles, and so has total area just over $7/36\approx 19.4\%$ of the full square.