It is a well-known result due to Novikov that the universal Pontryagin class of a tangent bundle over $\mathbb{Q}$ is a topological invariant.
Also, it is known that over $\mathbb{Z}$ Pontryagin class is not invariant.
What happens over $\mathbb{Z}_p?$
Pontryagin classes of manifolds with coefficients in $\mathbb{Z}_p$ are not topological invariants in general. In particular, by the argument in Section 4.4 of the book The Novikov Conjecture: Geometry and Algebra by Kreck and Lück, for any $n>1$, there exists a topological manifold $M$ with the following properties:
For any prime $p$ dividing $k$, it follows that the $p$-adic $n$th Pontryagin classes $p_n\in H^{4n}(M;\mathbb{Z}_p)$ for these two smooth structures are also different, since the map $H^{4n}(M;\mathbb{Z})\to H^{4n}(M;\mathbb{Z}_p)$ is nontrivial (explicitly, $H^{4n}(M;\mathbb{Z}_p)\cong \mathbb{Z}/(p^m)$ for the largest $m$ such that $p^m\mid k$, with the map $\mathbb{Z}/k\to\mathbb{Z}/(p^m)$ then being the quotient map).
Interestingly, this does not apply to every prime $p$. Indeed, the primes which can appear as factors of the number $k$ for some $n$ in this construction are $2$, odd primes $p$ such that the multiplicative order of $2$ mod $p$ is odd, and primes which divide the numerator of $\frac{B_{4n}}{4n}$ for some $n$ where $B_{4n}$ is the $4n$th Bernoulli number. This does not include all primes--for instance, this means $p=3$ will never be a factor of $k$. In fact, by the results of the paper Topologically invariant integral characteristic classes by Sharma, for each $n$ there is an integer $e_n$ which is a product of primes of the form above such that $e_np_n$ is a topological invariant. This implies that the Pontryagin classes with coefficients in any ring $R$ in which all such primes are units are topological invariants, which in particular includes $R=\mathbb{Z}_p$ for any prime $p$ not of this form.