To solve this problem, we need to analyze the properties of logarithmic and exponential functions, using the monotonicity of these functions to compare the sizes of $a$, $b$, and $c$.

First, let's analyze $a = \log_{0.3}2$. Note that $\log_{0.3}1 = 0$ and $\log_{0.3}\frac{1}{0.3} = -1$ since $0.3^{-1} = \frac{1}{0.3}$. Given that $\log_{0.3}2$ is a logarithm of a number between $\frac{1}{0.3}$ and $1$, and considering the logarithm base is less than 1, which makes the logarithm function decreasing, we have:
-1 = \log_{0.3}\frac{1}{0.3} 1$, therefore $b<-1$ which implies: b < a. Finally, analyze $c = 0.2^{0.3}$. We know that $0.2^{0} = 1$ and that 0.3 is positive, so the value of $c$ must be between 0 and 1 (since exponential functions are monotonically increasing when the base is between 0 and 1): 0 < 0.2^{0.3} < 0.2^{0} = 1. Thus, $c$ is between $0$ and $1$. Bringing it all together we have: b < -1 < a < 0 < c < 1.$$Hence, the relationship is:$$\boxed{b < a < c}.$$$$

Therefore, the correct answer is D.