Here is an explicit example where the answer is no. Start from the collection $K_p = \mathbf{Q}_p$ for every prime $p$. It obviously comes from the number field $\mathbf{Q}$. Now do the following strange thing: choose a non-empty set of primes $S$ of density 0, and replace $\mathbf{Q}_p$ by the unramified quadratic extension $\mathbf{Q}_{p^2}$ for each $p \in S$. If this new collection were to come from a number field $K$, then the set $\mathcal{P}$ of primes $p$ lying above some prime ideal $\mathfrak{p}$ in $K$ with inertia degree $f(\mathfrak{p}/p)=1$, would have density 1. Let $L$ be the Galois closure of $K$ and let $G=\mathrm{Gal}(L/\mathbf{Q})$ with its subgroup $H=\mathrm{Gal}(L/K)$. A prime $p$ which is unramified in $L$ belongs to $\mathcal{P}$ if and only if the conjugacy class of $\mathrm{Frob}_p$ in $G$ meets $H$. This means exactly that the action of $\mathrm{Frob}_p$ on $G/H$ has at least one fixed point. But a group acting transitively on a finite set of size $>1$ always contains an element which has no fixed point (a result due to Jordan I think, in any case it follows from the orbit-counting theorem). So if $n=[K:\mathbf{Q}]>1$ then the Cebotarev theorem for $L/\mathbf{Q}$ implies that $\mathcal{P}$ has density $<1$, a contradiction.
This example might look artificial, but the point is that (as noted in the comments) the Cebotarev theorem provides strong information about the distribution of splitting types of primes, at least in the case of Galois extensions of $\mathbf{Q}$ (it may be natural to consider this case first). But even having the correct densities is far from sufficient. As an example (taken from Serre's Lectures on $N_X(p)$, 3.4.4), take an elliptic curve $E$ over $\mathbf{Q}$ without complex multiplication and consider the condition $(A) : a_p(E)>0$ for primes $p$. It is expected that $(A)$ is independent of any condition of Cebotarev-type for number fields, and since the Sato-Tate conjecture is known for $E$, it should be possible to prove that a family of local fields modeled on condition $(A)$ cannot come from a number field. Actually, the condition $(A)$ is also natural in the sense that it is related to an automorphic form, but this is an automorphic form on $\mathrm{GL}_2$ and not $\mathrm{GL}_1$. And I guess there should be densities which are not naturally related to any automorphic form, since the set of automorphic forms is countable.
