1. Simplify the given product

We use the identity $\cos(60^\circ-\theta)\cos(60^\circ+\theta)=\cos^2 60^\circ-\sin^2\theta=\frac14-\sin^2\theta.$ But since the interval is in radians, it is cleaner to use the standard product identity with $\pi/3$:

$\cos\theta\cos\left(\frac\pi3-\theta\right)\cos\left(\frac\pi3+\theta\right).$

A standard identity is $4\cos x\cos\left(\frac\pi3+x\right)\cos\left(\frac\pi3-x\right)=\cos 3x.$

Hence, $\cos\theta\cos\left(\frac\pi3-\theta\right)\cos\left(\frac\pi3+\theta\right)=\frac14\cos 3\theta.$

So the given condition becomes $\left|\frac14\cos 3\theta\right|\le \frac18.$

Multiplying by $4$, $|\cos 3\theta|\le \frac12.$

2. Find the maximum possible value of $\cos 3\theta$ under this condition

Since $|\cos 3\theta|\le \frac12,$ we must have $-\frac12\le \cos 3\theta\le \frac12.$

Therefore, the maximum value attained by $\cos 3\theta$ is $\frac12.$

So we need all $\theta\in[0,2\pi]$ such that $\cos 3\theta=\frac12.$

3. Solve $\cos 3\theta=\frac12$

We know $\cos \phi=\frac12 \quad \Rightarrow \quad \phi=2n\pi\pm \frac\pi3.$ Thus, $3\theta=2n\pi\pm \frac\pi3.$

So, $\theta=\frac{2n\pi}{3}\pm \frac\pi9.$

Now list all solutions in $[0,2\pi]$.

Using $3\theta=\frac\pi3,\ \frac{5\pi}3,\ \frac{7\pi}3,\ \frac{11\pi}3,\ \frac{13\pi}3,\ \frac{17\pi}3$ within $[0,6\pi]$ (since $3\theta\in[0,6\pi]$), we get

$\theta=\frac\pi9,\ \frac{5\pi}9,\ \frac{7\pi}9,\ \frac{11\pi}9,\ \frac{13\pi}9,\ \frac{17\pi}9.$

These are all the required values in $[0,2\pi]$.

4. Find their sum

$\sum \theta=\frac\pi9(1+5+7+11+13+17).$

Now, $1+5+7+11+13+17=54.$

Hence, $\sum \theta=\frac{54\pi}{9}=6\pi.$

5. Choose the correct option

Therefore, the required sum is $6\pi.$

So the correct option is A.