No.
Suppose 2 of the vertices are (a,b) and (c,d), where a,b,c,d are integers. Then the 3rd vertex lies on the line defined by
(x,y) = 1/2 (a+c,b+d) + t ((d-b)/(c-a),-1) (t any real number)
and since the triangle is equilateral, we must have
||t ((d-b)/(c-a),-1)|| = sqrt(3)/2 ||(c,d)-(a,b)||
which yields t = +/- sqrt(3)/2 (c-a). Thus the 3rd vertex is
1/2 (a+c,b+d) +/- sqrt(3)/2 (d-b,a-c)
which must be irrational in at least one coordinate.