Here is an amalgamation of the comments of Ben Steinberg and Yves Cornulier with added details. I am grateful to Cornulier for his help over email.

The answer is *no*.

Suppose, for the sake of a contradiction, that the answer is yes. Then we may assign to every topologically finitely generated pro-p group a finitely generated nilpotent group. This gives a surjective map from the class of finitely generated nilpotent groups to the class of topologically finitely generated pro-$p$ nilpotent groups. Hence, the cardinality of the former is greater than that of the latter, and it follows that the class of topologically finitely generated pro-$p$ nilpotent groups is countable (since any finitely generated nilpotent group is finitely presented as it must be linear, by Hall’s embedding theorem). This conclusion, however, contradicts the following claim.

**Cornulier's claim**: There exists uncountably many topologically finitely generated pro-$p$ nilpotent groups.

**Proof**:
Consider the family of Lie algebras (147E) on page 61 from Ming-Peng Gong’s thesis, which are parametrized by the invariant $I(c) = \frac{(1-c+c^2)^3}{c^2(c-1)^2}$ where $c \in \mathbb{Z}_p$. Since $\mathbb{Z}_p$ is uncountable and $I(c)$ is finite to one, it follows that this is an uncountable family. By abusing notation a bit, we will simply denote this family by $\{ \mathfrak{g}_c : c \in \mathbb{Z}_p \}$.

By Ado’s theorem, we may realize any $\mathfrak{g}_c(\mathbb{Q}_p)$ as a Lie subalgebra of $\mathfrak{gl}(V)$, where $V$ is a vector space over $\mathbb{Q}_p$. Then by Engel’s theorem (also see Tao), we may embed $\mathfrak{g}_c(\mathbb{Q}_p)$ into an upper-triangulized lie algebra over $V$ (that is there exists a basis in $V$, where $\mathfrak{g}_c(\mathbb{Q}_p)$ is upper triangularized with respect to this basis). The Baker-Campbell-Hausdorff formula now applied to $\mathfrak{gl}(V)$ takes $\mathfrak{g}_c(\mathbb{Q}_p)$ to a Lie Group, $G_c(\mathbb{Q}_p)$. While this is not needed, note that the family $\{ \mathfrak{g}_c \}$ consists of Lie algebras that are of class 3, so the Baker-Campbell-Hausdorff formula giving multiplication simply as

$$
x*y=x+y+ \frac{1}{2}[x,y]+\frac{1}{12}([x,[x,y]]-[y,[x,y]]).
$$

It follows that we may actually define $G_c$ over $\mathbb{Z}_p$ (so long as $p > 3$) and $G_c(\mathbb{Z}_p)$ is compact. For these cases we see, somewhat concretely, how a Lie group $G_c (\mathbb{Q}_p)$ contains a compact subgroup that, roughly speaking, completely determines $G_c(\mathbb{Q}_p)$. For the general case, we set $K_c$ to be a maximal compact subgroup of $G_c(\mathbb{Q}_p)$. Since we are working p-adically, it turns out that $K_c$ is also open! And, as Cornulier stated in his comments below, any open subgroup of a Lie group shares the same Lie algebra as that of the bigger Lie group. Thus, $K_c$ and $G_c(\mathbb{Q}_p)$ share the same Lie algebra. It follows that distinct $c$ yield non-isomorphic $K_c$, and so the family $\{ K_c \}$ is uncountable.

Through the embedding of $G_c(\mathbb{Q}_p)$, we see that $K_c$ embeds into $U_n(\mathbb{Q}_p)$ (this is the set of upper triangular matrices with ones along the diagonal realized by the basis given above over $V$). Since $K_c$ is compact, only finitely many fractions can appear in this embedding into $U_n(\mathbb{Q}_p)$, thus we may find an automorphism of $U_n(\mathbb{Q}_p)$ so that the image of $K_c$ is in $U_n(\mathbb{Z}_p)$. It follows then that $K_c$ is pro-$p$. It is a well-known result of Lazard (his solution to Hilbert's fifth problem) that $p$-adic analytic and pro-$p$ groups are finite rank. **QED**

**tl;dr**: The take away from this is that nilpotent groups are very strictly determined by their nilpotent lie algebra, and doing things
$p$-adically just makes this correspondence stronger.