The answer is yes, and indeed, $Q$ is enough for Löb’s theorem:
Theorem. Let $T\supseteq Q$, and let $\tau\in\Sigma_1$ define an axiom set for $T$ in $\mathbb N$. Then
$$T\vdash(\Box_\tau\phi\to\phi)\implies T\vdash\phi$$
for all sentences $\phi$, where $\Box_\tau$ denotes the formalized provability predicate for $\tau$.
This was proved by Pudlák [1], even in a stronger form using restricted provability predicates (he states it for the second incompleteness theorem, but the argument for Löb’s theorem is essentially the same).
I will sketch the proof below. We rely on a few properties of Buss’s theory $S^1_2$ (see e.g. Chapter V in Hájek and Pudlák [2]): $S^1_2$ is a finitely axiomatizable fragment of arithmetic, it is interpretable in $Q$ on a definable cut, and it proves suitable versions of the usual Hilbert–Bernays–Löb derivability conditions.
Assume that $T\vdash\Box_\tau\phi\to\phi$, and let $I$ be an interpretation of $S^1_2$ in $Q$ on a definable cut. We have
$$T+\neg\phi\vdash\bigl(\neg\Box_\tau\phi\bigr)^I$$
as $\Pi_1$ statements are preserved downwards to cuts. Since $S^1_2$ is finitely axiomatizable, there exists a finite theory
$$\tag{$*$}U\subseteq T+\neg\phi$$
such that
$$\tag{$**$}U\vdash(S^1_2+\neg\Box_\tau\phi)^I.$$
We have the following derivability conditions for all sentences $\psi$ and $\chi$:
$$\begin{gather}
\tag{1}U\vdash\psi\implies S^1_2\vdash\Box_U\psi,\\
\tag{2}S^1_2\vdash\Box_U(\psi\to\chi)\to(\Box_U\psi\to\Box_U\chi),\\
\tag{3}S^1_2\vdash\Box_U\psi\to\Box_U(\Box_U\psi)^I.
\end{gather}$$
Using Gödel’s diagonal lemma, let $\nu$ be a sentence such that
$$\tag{$*{*}*$}S^1_2\vdash\nu\leftrightarrow\neg\Box_U\nu^I.$$
Then
$$\begin{align}
S^1_2\vdash\neg\nu
&\to\Box_U\nu^I&\text{by }&(*{*}*)\\
&\to\Box_U\bigl(\Box_U\nu^I\bigr)^I&\text{by }&(3)\\
&\to\Box_U\bigl(\Box_U\nu^I\to\neg\nu\bigr)^I&\text{by }&(*{*}*),(**),(1)\\
&\to\Box_U\neg\nu^I&\text{by }&(2)\\
&\to\Box_U\bot&\text{by }&(2)\\
&\to\Box_\tau\phi,
\end{align}$$
where the last step follows using the formalized deduction theorem from the fact that the axioms of $U$ consist of $\neg\phi$ and a finite list of axioms that satisfy $\tau$ (provably in $S^1_2$ as $\tau$ is $\Sigma_1$).
Thus,
$$\begin{align}
S^1_2+\neg\Box_\tau\phi&\vdash\nu,\\
U&\vdash\nu^I,&\text{by }&(**)\\
S^1_2&\vdash\Box_U\nu^I,&\text{by }&(1)\\
S^1_2&\vdash\neg\nu,&\text{by }&(*{*}*)\\
U&\vdash\neg\nu^I,&\text{by }&(**)\\
U&\vdash\bot,\\
T&\vdash\phi.
\end{align}$$
References
[1] Pavel Pudlák: Cuts, consistency statements and interpretations, Journal of Symbolic Logic 50 (1985), no. 2, pp. 423–441, DOI: 10.2307/2274231.
[2] Petr Hájek, Pavel Pudlák: Metamathematics of first-order arithmetic, Springer, 1994, 2nd ed. 1998, 3rd ed. Cambridge Univ. Press 2017.
