Can the CRT speed-up Paillier decryption by more than a factor of two?
Since CRT is an isomorphism computing $m_p=m\bmod q$ and $m_q=m\bmod q$ directly is possible. To see this in the formulas above replace the $\bmod n$ step with $p$ and $q$. 

To the question, I don't know if working in $\mathbb{Z}_{p^2}^{*}$ and $\mathbb{Z}_{q^2}^{*}$ could let you compute $m_p$ and $m_q$ more quickly than simply doing the steps modulo the prime factors of $n$. One thing that could help is to reduce $\lambda$ by the order of $\mathbb{Z}_{p^2}^{*}$ (or $q^2$ respectively). The order is given by the Euler totient function of $p^2$ which is $\phi(p^2) = p(p-1)$. This helps when $p$ or $q$ is small but in general doesn't speed things up. 

The only other comment I can make is the improvement you want would need to exploit some property of the group of elements of the form $x=c^\lambda\bmod n^2$ under multiplication. This is the group of elements of order dividing $n$. Its two nontrivial subgroups are the elements of order dividing $p$ and $q$.