There is a minor ambiguity in the term "Riemann integral": it tends to be used both for Riemann's original formulation -- which involves tagged partitions and requires convergence in a very strong sense: uniformly in the mesh (or norm) $||\mathcal{P}||$ of the partition $\mathcal{P}$ -- and also G. Darboux's later simplification in terms of upper and lower sums and upper and lower integrals, which is for most purposes technically easier to work with and thus is the one which is carefully developed in most undergraduate texts.
The ambiguity can be justified by the fact the Riemann and Darboux theories give different descriptions of what ultimately turns out to be the same linear functional: a function $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable if and only if it is Darboux integrable (the hard part of this is to show that Darboux integrable functions are Riemann integrable) and when these conditions hold the associated real number $\int_a^b f$ is the same. For a careful exposition of the Darboux and Riemann integrals including a comparison between the two, see Chapter 8 of these notes.
I bring up the distinction between Darboux and Riemann because it is relevant to your question of boundedness of integral functions, and because of the two helpful answers already left to this question, one addresses the Darboux case and the other the Riemann case. Either way though the following simple observation lies at the heart of the matter.
For $f: [a,b] \rightarrow \mathbb{R}$, the following are equivalent:
(i) $f$ is bounded above (respectively, bounded below).
(ii) For any partition $\mathcal{P} = \{a= x_0 < x_1 < \ldots < x_{n-1} < x_n = b\}$, the restriction of $f$ to each subinterval $[x_i,x_{i+1}]$ is bounded above (respectively, bounded below).
So if $f$ is unbounded above, then for any partition $\mathcal{P}$, there is at least one subinterval $[x_i,x_{i+1}]$ on which $f$ is unbounded above, hence the upper sum $\mathcal{U}(f,\mathcal{P})$ does not exist as a real number, so we can't even define the Darboux integral. Alternately, if we want to work in the extended real numbers, we would say that if $f$ is unbounded above, $\mathcal{U}(f,\mathcal{P}) = \infty$. Similarly, if $f$ is unbounded below, $\mathcal{L}(f,\mathcal{P}) = -\infty$ (c.f. Proposition 8.2 in the linked notes). This means: if $f$ is unbounded above then $\overline{\int}_a^b f = \infty$, and if $f$ is unbounded below then $\underline{\int}_a^b f = -\infty$. With this extended definition we would define a function to be Darboux integrable if and only if its upper and lower integrals are both finite and are equal, so we see that Darboux integrable functions are bounded.
For the Riemann integral there is a similar argument: if $f$ is unbounded above, then no matter what partition we choose, then for any $M > 0$ there will be a tagging $\tau$ -- i.e., a choice of sample point $x_i^* \in [x_i,x_{i+1}]$ such that the Riemann sum $R(f,\mathcal{P},\tau) = \sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$ is greater than $M$. This is a nice exercise: the idea is to use the above observation and choose one subinterval $[x_i,x_{i+1}]$ on which $f$ is unbounded above, choose the sample points in the other subintervals arbitrarily, and then choose $x_i^* \in [x_i,x_{i+1}]$ so that $f(x_i^*)$ is large enough to make the entire Riemann sum come out greater than $M$. Thus we see that if $f$ is unbounded above it cannot be Riemann integrable (but it's definitely a result, not a definition, in this case), and similarly if $f$ is unbounded below.
To address the rest of your question: yes, when one writes "Riemann integrable" one generally means to neglect the case of improper Riemann integrals, which of course can be finite even for some unbounded functions. This is true notwithstanding the fact that the same notation $\int_a^b f$ is used for Riemann integrals and improper Riemann integrals. Generally, anyway: in any particular case you should check to confirm that the terminology is being used in this way.
Added: The class of Riemann-Darboux integrable functions was characterized by Lebesgue (though my colleague Roy Smith has shown me a passage in Riemann's work showing that he had the result as well).
Theorem (Lebesgue Criterion) For a function $f: [a,b] \rightarrow \mathbb{R}$, the following are equivalent:
(i) $f$ is Riemann integrable.
(ii) $f$ is bounded, and the set of discontinuities of $f$ has measure zero.
Since this result was first published in the 20th century, one can correctly infer that it is not really needed in the development of the Riemann integral. From a pedagogical perspective, I would rather have students learn to use less heavy tools with more dexterity. Nevertheless I give a proof in $\S$ 8.5 of my notes which does not use any measure theory or even the theory of countable/uncountable sets. (The proof given there is not due to me; it is taken from notes of A.R. Schep.)
Note also that another answer to this question currently contains a false statement of this result. The characteristic function of the classical middle thirds Cantor set shows that "measure zero" cannot be replaced with "countable".
