Solution:

The answer is $6 < S \leq 10$. This can be shown by considering several cases, as shown in the 14 figures below.

Square 1

Square 8

Square 2

Square 9

Square 3

Square 10

Square 4

Square 11

Square 5

Square 12

Square 6

Square 13

Square 7

Square 14

Observe that in every case, there is either a horizontal or a vertical line segment of length $1$ drawn inside the square, as well as some additional segments. For example, there are three such vertical lines inside Square 1, two inside each of Squares 2-3, and one inside each of Squares 4-7. The figures in the first row are rotated by $90^{\circ}$ to give corresponding partitions of the square in the second row, each with a horizontal line cutting through the whole square; except for Square 14, which displays two lines, a horizontal and a vertical one, that pass through the center of the square.

The total perimeter $S$ of the four rectangles in each case is the original perimeter $4$ of the square, plus twice the length of all line segments drawn inside the square since each of these must be counted twice for the two (or more) rectangles that include them in their perimeters. So we must have $S = 4 + 2 \cdot 1 + 2a = 6 + 2a$, where $2 \cdot 1$ stands for twice a segment that crosses all the way through the square and $a$ stands for any additional segment(s) inside the square. Since $a > 0$ in all cases, we have $S > 6$.

The maximum of $S = 10$ is achieved by cutting the square into four $1 \times \frac{1}{4}$ rectangles with three parallel cuts, as in Squares 1 and 8.

All values in between can be achieved as well. Indeed, the total perimeters $S$ are written on top of each of the 14 cases. By shifting left or right the length $1$ vertical segments in Squares 2, 3, 4, or by shifting up or down the length $1$ horizontal segments in Squares 9, 10, 11, we can make the length $x$ vary from $0$ to $1$: $0 < x < 1$. Thus, for example, Square 4 can achieve any total perimeter $S = 6 + 4x$ between $6$ and $10$: $6 < S < 10$. In a similar way, Squares 5 and 12 can be drawn with lengths $x$ and $y$ varying between $0$ and $1$, and so these squares can also achieve any perimeter $S = 6 + 2x + 2y$ between $6$ and $10$.

Putting everything together, $S$ can be any number in the interval $(6, 10]$; i.e. $6 < S \leq 10$.