Yes, it is possible. First, if the parallelogram is not a rhombus, let $t > 1$ be the ratio of the long side to the short side. For any point in the plane, if the distance from that point to the origin lies within $[t^{2k}, t^{2k+1})$ (for any integer $k$), color it black; otherwise color it white. In this way, the two points of the parallelogram closest to the origin must be of different colors.

Next, consider the case where the parallelogram is a rhombus. Let its interior angle that is less than or equal to $90°$ have measure $2a$, i.e., $a \le 45°$. Then the angle between a side and a diagonal is either $a$ or $90° - a$. Now color as follows: measuring the elevation angle from the center with respect to the $x$-axis direction, divide the interval $[0°, 90°)$ into $[0, a)$, $[a, 2a)$, ..., $[ka, 90°)$, and color the points whose argument falls in the first interval black, the second white, alternating black and white in this manner. For the colors of angles beyond $90°$, follow this rule: if the arguments of two points differ by $90°$, then these two points are the same color. In this way, any three consecutive points (in counterclockwise order, labeled $A, B, C$) whose arguments differ successively by $a$ cannot all be the same color; in particular: when the elevation angles of $A, B$ lie in $[0, 90°)$, $A, B$ are of different colors; if the elevation angle of $A$ lies in $[0, 90°)$ but the elevation angles of $B, C$ lie in $[90°, 180°)$, then $B, C$ are of different colors. (There cannot be a situation where $A, B, C$ lie in three different quadrants, since $a \le 45°$.) All other situations are symmetric to these. Moreover, differing by $90° - a$ is equivalent to differing by $a$ with respect to the coloring, and thus the result is proven.