1. We need to evaluate $I=\int_0^\pi |\cos x|^3\,dx.$

2. On the interval $[0,\pi]$, the sign of $\cos x$ changes at $x=\frac{\pi}{2}$:

• For $0\le x\le \frac{\pi}{2}$, $\cos x\ge 0$, so $|\cos x|=\cos x$.
• For $\frac{\pi}{2}\le x\le \pi$, $\cos x\le 0$, so $|\cos x|=-\cos x$.

Thus, $I=\int_0^{\pi/2} \cos^3 x\,dx+\int_{\pi/2}^{\pi} (-\cos x)^3\,dx.$ Since $|\cos x|^3=(|\cos x|)^3$, it is easier to use symmetry: $|\cos(\pi-x)|=| -\cos x|=|\cos x|.$ So the function is symmetric about $x=\frac{\pi}{2}$, hence $I=2\int_0^{\pi/2} \cos^3 x\,dx.$

3. Now evaluate $\int_0^{\pi/2} \cos^3 x\,dx.$ Write $\cos^3 x=\cos x(1-\sin^2 x).$ Let $u=\sin x \quad\Rightarrow\quad du=\cos x\,dx.$ When $x=0$, $u=0$; when $x=\frac{\pi}{2}$, $u=1$. Therefore, $\int_0^{\pi/2} \cos^3 x\,dx=\int_0^1 (1-u^2)\,du.$

4. Compute the integral: $\int_0^1 (1-u^2)\,du=\left[u-\frac{u^3}{3}\right]_0^1=1-\frac13=\frac23.$

5. Hence, $I=2\cdot \frac23=\frac43.$

6. So the correct option is $\boxed{\text{A }\frac43}.$