Here is the full solution. The answer is, indeed, $5/2$. An example was already presented by the OP. Now we need to prove the inequality.
First of all we notice that for any color (blue or yellow) the sum of height/width (or width/height) ratios is always at least $1/2$.
Indeed, since all dimensions do not exceed $1$, we have (e.g., for blue color)
$$
\sum \frac{w_i}{h_i} \geqslant \sum w_i h_i = \frac 12 \,.
$$
as the final sum is the total area of all blue rectangles.
Second, we observe that either the blue rectangles connect the left and the right sides of the square, or the yellow rectangles connect the top and the bottom sides. We leave that as an exercise for the readers :) (Actually, as you will see below, it would suffice to show that either the sum of all blue widths or the sum of all yellow heights is at least $1$.)
Without loss of generality, assume that the blue rectangles connect the lateral sides of the large square. Then we intend to prove that
$$
\sum \frac{w_i}{h_i} \geqslant 2 \,,
$$
where the summation is done over the blue squares. Combining that with the inequality $\sum h_i/w_i \geqslant 1/2$ for the yellow squares we will have the required result, namely that the overall sum is always at least $5/2$.
Since the projections of the blue squares onto the bottom side must cover it completely, we have $\sum w_i \geqslant 1$. We also have $\sum w_ih_i = 1/2$. Now all we need is the following fact.
Lemma. Two finite sequences of positive numbers $\{w_i\}$ and $\{h_i\}$, i = $1$, ... , $n$ are such that
$$
\sum w_i = W, \qquad \sum w_ih_i = S \,.
$$
Then
$$
\sum \frac{w_i}{h_i} \geqslant \frac{W^2}S \,.
$$
Proof. We will use the well-known Jensen's inequality (which follows from the geometric convexity of the area above the graph of any convex function) for function $f(x) = 1/x$. That gives us
$$
\sum \frac{w_i}W f(h_i) \geqslant f \left( \sum \frac{w_i}W h_i \right) \,.
$$
In other words
$$
\frac1W \sum \frac{w_i}{h_i} \geqslant \frac1{\sum \frac{w_i}W h_i } =
\frac{W}{\sum w_i h_i} = \frac WS \,.
$$
and the required inequality immediately follows. $\square$
Applying this lemma to our case where $W \geqslant 1$ and $S = 1/2$ completes our solution.
Best Answer
I believe the answer is yes, there always exists a pair of sub-rectangles that do not overlap. Here is an outline proof.
Let the segment of the long cut that lies between the two points of intersection with the shorter cuts be called the Central Segment (CS).
Suppose that all four blue sub-rectangles overlap all four red sub-rectangles.
Then consider one blue sub-rectangle R. Each of the four red sub-rectangles must contain a point that is within R. Take four such points, one from each red sub-rectangle, and join them to make a quadrilateral Q.
Each edge of Q must lie within R (since it connects two points in R). Therefore Q lies fully within R.
Now Q must either intersect with the CS of the red rectangle, or it must fully enclose it (since the only way to draw such a quadrilateral without intersecting the red CS is to encircle it). Therefore the red CS passes through R.
The same is true for each blue sub-rectangle. Therefore the red CS passes through each blue sub-rectangle.
Let K and M be the two ends of the red CS, as shown in this diagram:
Note any segment that passes through each blue sub-rectangle - as the red CS does here - must cross both the blue CS and the two shorter cuts of the blue rectangle. (Such a segment may or may not continue beyond the bounds of the blue rectangle.)
Now consider the red sub-rectangle with a vertex at M whose edges do not overlap the red CS. There are two possible positions of this sub-rectangle, shown by the two dashed rectangles MOPL and MNQL in the diagram. But neither of these can possibly overlap with the blue rectangle that either contains or is nearest to K (the other end of the red CS), shown as GDFH in the diagram. This can be proved by (for example) considering the projection of the rectangle GDFH onto the line KL: this projection is entirely within the segment KM, while the projections of the potential red sub-rectangles MOPL and MNQL are the segment ML.
Therefore we have a red sub-rectangle that doesn't overlap with a blue sub-rectangle, and have reached a contradiction.