I came up with a positive proof for the title's question; however, it seems very counter-intuitive to me because the set of binary sequences seems much smaller, yet is uncountable. Moreover, we can forgo positivity as a condition, but I won't for simplicity.
Perhaps I have made an elementary mistake?
Proof.
Suppose $q_n$ is a positive sequence of rationals diverging to infinity. Since it diverges there are only finitely many $q_n$ below any given $i\in\mathbb{N}$.
The set of positive rationals below $i$, $Q_i$, is countable.
The set of orderings of any finite subset of $Q_i$ is finite; in turn, for any finite subset of $Q_i$, the set of all its finite sequences is finite. The number of subsets of $Q_i$ of a given size is countable; as is the number of sizes. In turn, the set of all ordered finite subsets of $Q_i$, $\tilde{Q}_i$ is countable.
Then the set $\bigcup_{i\in\mathbb{N}} \tilde{Q}_i$ is countable, and the set of positive rational sequences diverging to infinity injects into it.
Best Answer
I don't understand how the last sentence of your argument is supposed to work -- how do you clam that the set you're talking about injects into $\bigcup_i \bar Q_i$? -- but what you're trying to prove is certainly a falsehood.
You know that there are uncountably many infinite binary sequences; however, if $(b_n)$ is a infinite binary sequence, then $$ c_n = n + b_n $$ is an infinite rational sequence that diverges to $+\infty$, and the mapping from $(b_n)$s to $(c_n)$s is injective, so even just there you have uncountably many different rational sequences that diverge to infinity.