Use the condition that $x=2$ is a root of $x^2+px+q=0$.

Since $2$ is a root, $2^2+2p+q=0$ $4+2p+q=0$ $q=-2p-4$

Simplify the expression inside cosine.

We have $x^2-4px+q^2+8q+16$

Now, $q^2+8q+16=(q+4)^2$

Since $q=-2p-4$, $q+4=-2p$ $q^2+8q+16=4p^2$

Therefore, $x^2-4px+q^2+8q+16=x^2-4px+4p^2=(x-2p)^2$

Rewrite $f(x)$.

For $x\ne 2p$, $f(x)=\frac{1-\cos\big((x-2p)^2\big)}{(x-2p)^4}$

Evaluate the limit of $f(x)$ as $x\to 2p^+$.

Let $t=(x-2p)^2$ Then as $x\to 2p^+$, we have $t\to 0^+$.

So, $f(x)=\frac{1-\cos t}{t^2}$

Using the standard expansion, $1-\cos t \sim \frac{t^2}{2} \quad (t\to 0)$

Hence, $\lim_{x\to 2p^+} f(x)=\lim_{t\to 0^+}\frac{1-\cos t}{t^2}=\frac12$

Now evaluate the greatest integer.

Since $f(x)\to \dfrac12$ as $x\to 2p^+$, and for $x$ sufficiently close to $2p$ from the right, $f(x)$ remains in $(0,1)$, we get $[f(x)]=0$ for all such $x$.

Therefore, $\lim_{x\to 2p^+}[f(x)]=0$

Check options.

The correct option is: $\boxed{\text{C: }0}$