First note that it is not true that regularity above $\mathfrak q$ implies regularity of the fibre.
For example, consider the map $\mathbb k[s] \to \mathbb k[t]$ given by $s \mapsto t^2$.
Each prime in $k[t]$ is regular, and so in particular each prime of $k[t]$ above
the prime $(s)$ is reqular. (In fact there is just one of them, namely $(t)$.) On the other hand, the fibre over $(s)$ is the ring $k[t]/(t^2)$, which is a non-regular
If you want an arithmetic example instead, consider the inclusion
$\mathbb Z \subset \mathbb Z[i]$, and take the prime $\mathfrak q = (2)$ downstairs,
which has a unique prime $(1+i)$ lying over it. Again, every prime in the PID $\mathbb Z[i]$
is regular, but the fibre $\mathbb Z[i]/2 = \mathbb Z[i]/(1+i)^2$ is a non-regular local
(The general phenomenon is that a map $X \to Y$ between smooth spaces can have
non-smooth fibres: these occur at the critical points of the map.)
As for your question, your asking if the fact that a map $X\to Y$ has
a smooth fibres over some point implies that the target is smooth at that point.
This is also false in general.
Consider for example the identity map
$k[t^2,t^3] \to k[t^2,t^3]$. The fibre over $(t^2,t^3)$ is just $k$, which is a regular local ring. But $K[t^2,t^3]$ is not regular at $(t^2,t^3).$ (Of course this example is cheap, but its existence foreshadows the existence of many other counterexamples, for example for any etale map $k[t^2,t^3] \to R$, of which there are many non-trivial examples,
as well as my trivial example.)
On the other hand, when the base is Spec $\mathbb Z$, which is very nicely behaved
(regular, Noetherian, excellent, perfect residue fields, ... ), if the map
$X \to $ Spec $\mathbb Z$ is flat and of finite type (e.g. arising from an inclusion
$\mathbb Z \subset R$ of the form you envisage) then having regular (and hence smooth) fibre
at one point implies being smooth in a neighbourhood of that point, and a smooth map over
a regular base has a regular total space --- thus $X$ will be regular in a neighbourhood of the regular fibre. In particular, the points of the regular fibre will themselves be regular in $X$.
In particular, in your special case $\mathbb Z \subset R$, the answer is "yes".
Edit: I should note that in your situation, where $R$ is a subring of a number field,
this "yes" is easily proved directly: one combines the fact that $R/\mathfrak q$ is
regular with the fact that it is a priori finite (in cardinality) to see that
it is a product of finite field extensions of $\mathbb Z/\mathfrak q$, and hence
that the completion of $R$ at $\mathfrak q$ is a product of DVRs, and hence that
$R$ is a DVR --- and thus regular --- after localization at each prime above $\mathfrak q$.
The point of the more highbrow explanation above is to indicate how one thinks about
such questions geometrically --- which is normally the easiest way to see what should
be true and what should be false for these kinds of questions.