Does $\mathcal{P}\neq\mathcal{NP}$ imply $\mathcal{NP}\neq\mathcal{NP}^{\mathcal{NP}}$?
No, this implication is not known to be true, however it is believed that those are all separate classes.  (Unfortunately, I don't have any references to back up my claim now, but I still post this as an answer to point out the following.)

You are asking whether $\text{P} \neq \text{NP}$ implies that the polynomial hierarchy $(\text{PH})$ does not collapse to its first level.

The class $\text{NP}^\text{NP}$ is also known as $\Sigma_2^P$ which is one of the classes on the second level of the polynomial hierarchy.