1. Start by simplifying the integrand:

$$(|\sin x|+|\cos x|)^2 = \sin^2 x + \cos^2 x + 2|\sin x\cos x| = 1 + 2|\sin x\cos x|.$$

Using $2\sin x\cos x = \sin 2x$, we get

$$2|\sin x\cos x| = |\sin 2x|.$$

So the integral becomes

$$I = \int_0^{20\pi} (1+|\sin 2x|)\,dx.$$

2. Split the integral:

$$I = \int_0^{20\pi} 1\,dx + \int_0^{20\pi} |\sin 2x|\,dx.$$

Thus,

$$I = 20\pi + \int_0^{20\pi} |\sin 2x|\,dx.$$

3. Evaluate $\int_0^{20\pi} |\sin 2x|\,dx$.

Let

$$u = 2x \quad \Rightarrow \quad d\nu = 2dx, \quad dx = \frac{d\nu}{2}.$$

When $x=0$, $\nu=0$; when $x=20\pi$, $\nu=40\pi$.

So,

$$\int_0^{20\pi} |\sin 2x|\,dx = \frac12 \int_0^{40\pi} |\sin \nu|\,d\nu.$$

Now, over one period $[0,2\pi]$,

$$\int_0^{2\pi} |\sin \nu|\,d\nu = 4.$$

Since $40\pi = 20(2\pi)$, there are $20$ full periods. Hence,

$$\int_0^{40\pi} |\sin \nu|\,d\nu = 20 \cdot 4 = 80.$$

Therefore,

$$\int_0^{20\pi} |\sin 2x|\,dx = \frac12 \cdot 80 = 40.$$

4. Substitute back:

$$I = 20\pi + 40 = 20(\pi+2).$$

5. Compare with the options:

$$\boxed{20(\pi+2)}$$

So the correct option is D.