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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.
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