Solution:

Since $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers whose squares sum to $25,$ the pair $(\lfloor x \rfloor, \lfloor y \rfloor)$ is one of the $12$ pairs $(\pm 5, 0),$ $(0, \pm 5),$ $(\pm 3, \pm 4),$ $(\pm 4, \pm 3).$ So $\mathcal{R}$ is the union of the $12$ unit squares whose lower-left corners are these points.

The map $(x, y) \mapsto (1 - x, 1 - y)$ permutes these squares, so $\mathcal{R}$ is symmetric under $180^\circ$ rotation about $Q = \left(\frac{1}{2}, \frac{1}{2}\right).$ The smallest enclosing disk is unique, so its center must be $Q.$ The farthest points of $\mathcal{R}$ from $Q$ are square corners such as $A = (4, 5)$ and $B = (5, 4),$ at distance $$\sqrt{\left(\tfrac{9}{2}\right)^2 + \left(\tfrac{7}{2}\right)^2} = \frac{\sqrt{130}}{2};$$ checking all twelve squares confirms no corner is farther.

Hence the minimum radius is $r = \frac{\sqrt{130}}{2},$ and $m + n = 130 + 2 = 132.$