For $$I(x) = \int \frac{\sec^2 x - 2022}{\sin^{2022} x} dx$$, if $$I\left(\frac{\pi}{4}\right) = 2^{1011}$$, then

We have $$I(x) = \displaystyle\int \dfrac{\sec^2 x - 2022}{\sin^{2022} x}\, dx$$ with $$I\left(\dfrac{\pi}{4}\right) = 2^{1011}$$.

We rewrite the integrand as $$\dfrac{\sec^2 x}{\sin^{2022} x} - \dfrac{2022}{\sin^{2022} x}$$. Now $$\dfrac{\sec^2 x}{\sin^{2022} x} = \dfrac{1}{\cos^2 x \cdot \sin^{2022} x}$$. We can write this as $$\dfrac{1}{\sin^{2022} x \cdot \cos^2 x}$$.

We try to recognize this as a derivative. Consider $$\dfrac{d}{dx}\left(\dfrac{\tan x}{\sin^{2022} x}\right)$$. Using the quotient rule: $$\dfrac{\sec^2 x \cdot \sin^{2022} x - \tan x \cdot 2022 \sin^{2021} x \cos x}{\sin^{4044} x}$$ $$= \dfrac{\sec^2 x}{\sin^{2022} x} - \dfrac{2022 \tan x \cos x}{\sin^{2022} x}$$ $$= \dfrac{\sec^2 x}{\sin^{2022} x} - \dfrac{2022}{\sin^{2022} x}$$

This is exactly our integrand! So $$I(x) = \dfrac{\tan x}{\sin^{2022} x} + C$$.

Now using $$I\left(\dfrac{\pi}{4}\right) = 2^{1011}$$: $$\dfrac{\tan(\pi/4)}{\sin^{2022}(\pi/4)} + C = \dfrac{1}{(1/\sqrt{2})^{2022}} + C = 2^{1011} + C = 2^{1011}$$, so $$C = 0$$.

Therefore $$I(x) = \dfrac{\tan x}{\sin^{2022} x}$$.

Now we compute $$I\left(\dfrac{\pi}{3}\right) = \dfrac{\tan(\pi/3)}{\sin^{2022}(\pi/3)} = \dfrac{\sqrt{3}}{(\sqrt{3}/2)^{2022}} = \dfrac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} = \dfrac{2^{2022}}{3^{1011} \cdot 3^{-1/2}} = \dfrac{2^{2022} \cdot 3^{1/2}}{3^{1011}} = \dfrac{2^{2022}}{3^{2021/2}}$$.

Let us be more careful. $$\sin(\pi/3) = \dfrac{\sqrt{3}}{2}$$, so $$\sin^{2022}(\pi/3) = \left(\dfrac{\sqrt{3}}{2}\right)^{2022} = \dfrac{3^{1011}}{2^{2022}}$$. Thus $$I(\pi/3) = \dfrac{\sqrt{3} \cdot 2^{2022}}{3^{1011}}$$.

Similarly, $$I\left(\dfrac{\pi}{6}\right) = \dfrac{\tan(\pi/6)}{\sin^{2022}(\pi/6)} = \dfrac{1/\sqrt{3}}{(1/2)^{2022}} = \dfrac{2^{2022}}{\sqrt{3}}$$.

Now we check Option A: $$3^{1010} I(\pi/3) - I(\pi/6) = 3^{1010} \cdot \dfrac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} - \dfrac{2^{2022}}{\sqrt{3}} = \dfrac{\sqrt{3} \cdot 2^{2022}}{3} - \dfrac{2^{2022}}{\sqrt{3}} = \dfrac{2^{2022}}{\sqrt{3}} - \dfrac{2^{2022}}{\sqrt{3}} = 0$$.

This confirms Option A.

Hence, the correct answer is Option A: $$3^{1010} I\left(\dfrac{\pi}{3}\right) - I\left(\dfrac{\pi}{6}\right) = 0$$.