Well, wouldn't it actually take fewer in most cases?
There are two basic arrangements of overlapping circles that, between them, encompasses a solid area greater than that of each individual circle. The orthogonal arrangement is easier to visualize especially relating to a rectangle, so we'll use it. One iteration of this arrangement of circles takes five of them; four in a square with their edges touching, and one in the middle filling the gap in the center.
One iteration would completely cover a square 2R on a side (it can be no wider than the center circle, of diameter 2R, or some of the square would be outside the addjacent points of the outer four circles). If you needed to cover a rectangle composed of two of these squares, 2R*4R, you'd need two iterations of the pattern, but (here's the kicker) two of the circles from each iteration are congruent with one other, and are redundant; 5 circles of radius R cover a 2R*2R square, but only 7 are needed for 2R*4R, not 10. Double the area again, to a square 4R*4R, and you need 4 iterations of the original shape (2 of the rectangular 7-circle pattern), but now all four iterations have a circle that is congruent with one from the other three, as well as another circle congruent with one other iteration; that's 7 circles out of 20 that are redundant so you only need 13.
The number of orthogonally-patterned circles of radius r that are required to cover a rectangle of area n*m where n and m are even multiples of r is the sum of two related products of n and m; to cover each dimension of the rectangle with circles, you need one more circle per dimension than half the number of radii in the dimension, so the first set of orthogonal circles is the product of those two quantities. Then, the second set fills in the gaps, and requires the product of half the number of circles as there are radii in each dimension. Thus, $N(r,n,m) = (n/2r+1)(m/2r+1) + nm/4r$. Halve the radius and you find that $N(r/2,n,m) = (n/r+1)(m/r+1) + nm/2r$, and if you instead double the dimensions you'll find that $N(r/2,n,m) = N(r,2n,2m)$.
Working backwards with this formula, a rectangle that requires 25 circles of radius r would actually be a square 6r on a side; $N(r, 6r, 6r) = (6r/2r + 1)(6r/2r + 1) + (6r/2r)(6r/2r) = 4^2 + 3^2 = 25$. Halving the radius of the circles, we'd need $N(r/2, 6r, 6r) = N(r,12r,12r) = 7^2 + 6^2 = 85$. That's one extreme; the other is a rectangle (2r*16r), using a variation of the coverage pattern where one circle covers each short end and the remainder of the shape is filled in with the normal pattern. The number of circles in that case reduces to the very simple equation $N(r,2r,2xr) = 3x+1$, which when solved for x=8 is 25. Halve r, and the most efficient pattern goes back to the one stated above, so $N(r/2, 2r, 16r) = N(r,4r, 32r) = 3*9 + 2*16 = 59$.
So, while it would take more than double the number of circles when the radius is halved, it would never take 4x the circles. So, the answer to the question as stated is "true"; a rectangle that requires 25 discs of radius $r$ to cover with overlap can be covered with 100 discs of radius $r/2$; in fact you could cover a larger rectangle with this many.