To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:

$$ \cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2 $$

We can then expand and simplify the integral using this formula. Let's proceed with this:

$$ \int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x $$

Now, let's expand the integrand and then integrate term by term:

$$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x $$

For the $\cos^2(2x)$ term, we again use the power reduction formula:

$$ \cos^2(2x) = \frac{1 + \cos(4x)}{2} $$

Let's substitute this into the integral and continue:

$$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x $$

Simplify and integrate:

$$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $$

$$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $$

$$ = \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}} $$

Evaluating this from $0$ to $\frac{\pi}{3}$:

$$ = \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right) $$

$$ = \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right) $$

$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$:

$$ = \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2} $$

$$ = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} $$

Now, combining terms we get the final result:

$$ = \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64} $$

$$ = \frac{\pi}{8} + \frac{7\sqrt{3}}{64} $$

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$:

$$ a = \frac{1}{8}, \quad b = \frac{7}{64} $$

Now we find $9a + 8b$:

$$ 9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} $$

$$ = \frac{9}{8} + \frac{7}{8} $$

$$ = \frac{16}{8} $$

$$ = 2 $$

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.