Okay, here's a non-trivial answer. I believe I can get arbitrarily close to a score of 1/sqrt(3) = 0.57735... .

Pick any natural number n, and let x = sqrt(3)/2. Using the theory of continued fractions one can find natural numbers p and q such that q > n and x < p/q < x + 1/(4sqrt(3)q^2). Now let our covering shape be a right angled triangle with sides 1/(2p) and x/p. The area is x/(4p^2). Clearly 2p^2 such triangles can cover the equilateral triangle exactly.

Now consider how many we can fit in the square. By using two of our covering triangle we can make a rectangle of sides 1/(2p) and x/p. By arranging such rectangles in a 2p by q grid we can perfectly tile a big rectangle with sides 1 and qx/p. Since x < p/q we have qx/p < 1, so this rectangle fits into the square.

The remaining area is 1 - qx/p, and since p/q < x + 1/(4sqrt(3)q^2) we have that this area is less than 1/(4sqrt(3)pq). Comparing this to the area of our covering triangle we find that our score is at most (p/q)(1/x)(1/sqrt(3)) which is less than (x + 1/(4sqrt(3)q^2))(1/x)(1/sqrt(3)) which is less than (1+1/(4sqrt(3)xn^2))(1/sqrt(3)). Since n was arbitrary this can come arbitrarily close to a score of 1/sqrt(3).

EDIT: I made a picture of the tiling of the square when p=13, q=15. The uncovered area is the tiny red line at the top. The score is 0.57756... .