The answer to a sharper question involving integers, rather than rationals, is affirmative.
Let $\lambda$ be a positive real algebraic integer that is greater in absolute value than all its Galois conjugates ("Perron number" or "PF number"). Then $\lambda$ is the Perron–Frobenius eigenvalue of a positive integer matrix.
(The converse statement is an integer version of the Perron–Frobenius theorem, and is easy to prove.)
In a slightly weaker form (aperiodic non-negative matrix), this is theorem of Douglas Lind, from
The entropies of topological Markov shifts and a related class of algebraic integers.
Ergodic Theory Dynam. Systems 4 (1984), no. 2, 283--300 (MR)
I don't have a good reference for the strong form, but it was discussed at Thurston seminar in 2008-2009. One interesting thing to note is that, while the proof can be made constructive, it is non-uniform: the size of the matrix can be arbitrarily large compared to the degree of $\lambda$.
The information you have does not determine the dominant eigenvector.
Let $G$ be the graph with vertex set $\{0,1,\ldots,7\}$ and adjacency matrix
$$
\left(\begin{array}{rrrrrrrr}
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right)
$$
Construct a second graph $H_2$ by joining a new vertex to the vertex 2, and a third
graph $H5$ by joining a new vertex to vertex 5. Then
$$
(A(H_2)^k e)_2 = (A(H_5)^k e)_5
$$
for all $k$. (For $k=0,\ldots,8$ the actual numbers are
$$
1,\\ 4,\\ 8,\\ 25,\\ 57,\\ 163,\\ 392,\\ 1073,\\ 2656)
$$
The Perron vectors are
$$
(1, 1, 1.579071, 1.460275, 1.019079, 1.168003, 0.5330099, 0.206667, 0.612263)
$$
for $H_2$ and, for $H_5$,
$$
(1, 1, 1.579071, 2.0725388, 1.631342, 2.134811, 0.974205, 0.377735, 0.827744)
$$
If you want positive matrices, take the sixth powers of $A(H_2)$ and $A(H_5)$. The relevant
property of $G$ is that the graphs $G\setminus2$ and $G\setminus5$ are cospectral, and their
complements are cospectral too.
All computations carried out in sage.
In a sense the problem is that you are getting a bit of information about each eigenspace,
whereas you want detailed information about a particular eigenspace.
Best Answer
Beware of Wikipedia! It is true that the infinite dimensional setting makes things slightly more delicate, but actually not so much.
Assuming that the positive cone $C\subset X$ under consideration is solid (i-e has non empty interior) and that your operator $T:X\to X$ is compact and strongly positive (i-e maps the positive cone $C$ into its interior $\overset{\circ}{C}$), then the following stronger conclusion holds: the spectral radius is a simple eigenvalue associated with a strictly positive eigenvector $v\in \overset{\circ}{C}$, and there is no other eigenvalue associated with (non necessarily strictly) positive eigenvectors.
You can find an elementary proof here (theorem 1.2)