Hilbert's axioms predate the development (or at least the wide adoption) of fully symbolic logic, so they are expressed in partially informal language -- though Hilbert strove to make them as precise as he could.
They include the Axiom of Archimedes, formulated in language that presupposes that the natural numbers are already known. As such, if we want to formalize the system in a modern sense, we'd need to add in some machinery for talking about integers, and it would be somewhat hard to avoid making that machinery powerful enough that Gödel's theorem would apply to it.
Furthermore, there's also an Axiom of Completeness which attempts to claim directly that the structure we're speaking about is maximal in a certain technical sense. In the modern view of formal theories, this hardly qualifies as an "axiom" at all, because it doesn't assert the truth of any formula interpreted inside the language of the theory itself. In modern terms, it appears to try to say that every existential formula of a certain shape that is true in some model of the other axioms must itself be elevated to axiomatic status.
However, it is not clear that "formulas that are true in some model of the other axioms" are even recursively enumerable -- which means that Hilbert's axioms, interpreted in this way, are not effective. In other words there's no systematic way to check whether a purported proof is valid or not. Therefore Gödel's incompleteness theorem does not apply to Hilbert's axioms. It seems at least plausible that if we interpret them inside set theory in the above sense, they do have $\mathbb R^3$ as their only model up to isomorphism. (That is, whatever the set theory in question considers $\mathbb R^3$ to be).
Tarski's axioms for geometry were created after formal logic was better developed. They form a genuine, effectively axiomatized, first-order theory, with no ad-hoc appeals to integers, sets, or model theory. They manage to be a complete theory because they are not strong enough to express or simulate arithmetic.
The price paid for the completeness in Tarski's case is that the language the axioms are formulated in is not expressive enough to speak of even finite sets of points, or for example general polygons. Every theorem has to be proved separately for triangles, quadrilaterals, pentagons, and so forth.
This restriction is unavoidable for a complete theory, because as soon as we extend the language with a way to speak of finite sets of points and lines in a reasonable way, we can use those sets as proxies for natural numbers (each set representing the number of points in it) and begin to speak about enough arithmetic that Gödel's incompleteness theorem will apply to it.
Later, Tarski gave his own axiomatization of Euclidean geometry that is entirely in first order logic so by Gödel's completeness theorem, it can demonstrate its consistency, it is decidable and complete.
As mentioned in the comments, this is a complete misunderstanding of what the completeness theorem says. Tarski's geometry cannot even speak about its own consistency, much less prove it.
The key point is that "proof" requires you to at least understand what theory you are assuming.
Both the Riemann Hypothesis and "$T$ is consistent" (where $T$ is any recursively enumerable theory) are $\Pi^0_1$ statements. This means that if there is a counterexample, then the statement is disprovable. Or in other words, if the statement is not disprovable, then it is true.
But what does "true" mean? In the context of arithmetic true means "true in the natural numbers". In other contexts, it is usually meaningless, or synonymous for "provable".
Yes. $\sf PA$ is consistent is a $\Pi^0_1$ statement, but we also can prove that it is a true statement, since $\Bbb N$ exists and it is a model of $\sf PA$. Or at least that is the case if your metatheory is sufficiently strong (e.g. set theory à la $\sf ZFC$). The point is that $\sf PA$ itself cannot prove that $\sf PA$ is consistent, and that $\sf ZFC$ cannot prove that $\sf ZFC$ is consistent.
The difference between that and the Riemann Hypothesis is that you don't really talk about consistency of a theory. So there is no natural candidate of a theory which is insufficient for proving the Riemann Hypothesis.
It is possible for a $\Pi^0_1$ statement to turn out equivalent to "$\sf ZFC$ is consistent", in which case it cannot be proved. But it would be a mind shattering surprise to modern mathematics if RH turns out to be such statement. So what we expect is that RH is either outright refutable from a weak theory like $\sf PA$, and if not then it is at least true in $\Bbb N$, which is "the thing we care about" for the Riemann Hypothesis.
Best Answer
Yes. That is exactly what it means. Consistency assumptions are axioms.
This gives rise to a natural hierarchy of axioms, specifically part of set theory, called large cardinal axioms which are stronger and stronger in consistency strength, and generally each one implies the weaker are consistent (and much more).
For example, the standard set theory, ZFC (Zermelo–Fraenkel with Choice) does not prove its own consistency strength, but we can add an axiom stating that it is in fact consistent. To that you can add an axiom that it is not only that ZFC is consistent, but "ZFC+ZFC is consistent" is also consistent. This can go on for a while.
But you can also just say that there are inaccessible cardinals, whatever they might be. This implies that ZFC is consistent, and much more. You can move to stating that there exists a weakly compact cardinal, which then implies that not only it is consistent that there is an inaccessible cardinals, but that it is consistent that every set is smaller in size from some inaccessible cardinal.
And the list continues.
Interestingly, though, while the large cardinal axioms are stating that some particular sets exists (or don't exist), consistency statements can be seen as axioms added to the theory of the natural numbers. So you can also investigate them from arithmetic theories such as PA or PRA, both of which are vastly weaker than ZFC.