Solution:

The $12$ segments split into the $4$ interior segments forming the central cross and $8$ boundary segments, and each unit square has exactly two interior sides (its two cross arms) and two boundary sides. Color the cross first. A square that already has $j$ red interior sides needs $2 - j$ red boundary sides, which can be chosen in $\binom{2}{2-j}$ ways: $1$ way if $j = 0$ or $j = 2,$ and $2$ ways if $j = 1.$

Group the $2^4 = 16$ cross colorings by the set of red arms. If all four arms have the same color ($2$ colorings), every square has $j = 0$ or $j = 2,$ contributing $1$ each: total $2.$ If exactly one arm is red or exactly one is blue ($8$ colorings), the two squares touching the odd arm have $j = 1$ and the others do not, contributing $2 \cdot 2 = 4$ each: total $32.$ If two adjacent arms are red ($4$ colorings), the squares have $j = 2, 1, 1, 0,$ contributing $4$ each: total $16.$ If two opposite arms are red ($2$ colorings), all four squares have $j = 1,$ contributing $2^4 = 16$ each: total $32.$

The number of colorings is $2 + 32 + 16 + 32 = 82.$