There are really two separate assertions here: one concerns the generic fiber $A = \mathbf{Q} \otimes_{\mathbf{Z}} R$, and the other concerns orders in semisimple $\mathbf{Q}$-algebras. To clarify the framework for such arguments (e.g., to convey how "characteristic zero" is a red herring and one can make arguments that work in all characteristics uniformly), we pose the assertions more generally:
(i) If $A$ is a nonzero finite-dimensional associative algebra over a field $k$ (e.g., $k = \mathbf{Q}$) and $A_K := K \otimes_{k} A$ is semisimple for one algebraically closed extension $K/k$ then the same holds for all algebraically closed extensions of $k$. (This is called "absolute semisimplicity'' over $k$.)
(ii) If $B$ is a noetherian domain (e.g., $\mathbf{Z}$) with fraction field $k$, $A$ is a nonzero finite-dimensional absolutely semisimple $k$-algebra, and $R$ is a $B$-order in $A$ (i.e., a $B$-subalgebra finitely generated as a $B$-module that spans $A$ as a $k$-vector space) then there exists a nonzero $b \in B$ such that $R[1/b]$ is an Azumaya algebra over $B[1/b]$ (i.e., $R[1/b]$ is finite flat as a $B[1/b]$-module and $R[1/b] \otimes_{B[1/b]} \kappa(\mathfrak{p})$ is absolutely semisimple over $\kappa(\mathfrak{p}) := {\rm{Frac}}(B[1/b]/\mathfrak{p})$ for all primes $\mathfrak{p} \in {\rm{Spec}}(B[1/b])$.
Each of these is a consequence of standard "direct limit" methods that are exhaustively developed and used in a vast array of situations in EGA IV$_3$--IV$_4$, combined with knowledge of the structure of nonzero finite-dimensional semisimple algebras over algebraically closed fields.
At a basic level this amounts to apt use of the Nullstellensatz and denominator-chasing, but is often much deeper than that because for properties such as flatness and properness and conditions on geometric fibers that are not expressed in terms of "equations" one has to come up with some substantial ideas.
Proof of (i): To save notation, we'll write $V_K$ to denote $V \otimes_k K$ for any $k$-vector space $V$. The center $Z$ of $A$ is the kernel of the system of linear conditions $a \mapsto ae_i - e_i a$ for a $k$-basis $\{e_i\}$ of $A$, so it is clear that $Z_K$ is the center of $A_K$ for any extension field $K/k$. By hypothesis there is some such algebraically closed $K$ for which $A_K$ is semisimple, and we want to deduce the same for all such $K/k$.
Fix an algebraic closure $\overline{k}$ of $k$, so there is a $k$-embedding of $\overline{k}$ into any such $K/k$. Hence, by replacing $k$ with $\overline{k}$ we are reduced to showing that if $k$ is algebraically closed and $K/k$ is an algebraically closed extension then $A_K$ is semisimple if and only if $A$ is semisimple. (This is an instance of what EGA called "the principle of algebraically closed fields", essentially a glorified version of the Nullstellensatz.)
We know that a (nonzero) finite-dimensional associative algebra over an algebraically closed field is semisimple if and only if it is a product of matrix algebras, in which case its center is a product of copies of the ground field. But over an algebraically closed field, the products of finitely many copies of the field are precisely the etale commutative algebras, and the property of being etale for a commutative algebra with finite linear dimension over a field is insensitive to extension of the ground field (consider the non-vanishing or not of the discriminant attached to a choice of basis). Hence, if either $A$ or $A_K$ is semisimple then each of $Z$ and $Z_K$ is etale (over $k$ and $K$ respectively), so we may now assume this etaleness holds.
The $k$-algebra $Z$ is now a direct product of copies of $k$. Since a direct product of two $k$-algebras is semisimple if and only if each factor is semisimple (and likewise over $K$), we may decompose $A$ as a direct product according to the primitive idempotents of $Z$ to reduce to the case $Z=k$. Then $Z_K=K$, so our task has become proving that $A$ is a matrix algebra if and only if $A_K$ is a matrix algebra. In either case $\dim_k A = n^2$ is a perfect square, so we may assume this dimension expression holds for some $n$.
Now comes the part with algebro-geometric content.
Consider the finite type affine $k$-scheme
$I$ representing the functor on $k$-algebras $R \rightsquigarrow {\rm{Isom}}(A_R, {\rm{Mat}}_n(R))$; it is easy to build $I$ by choosing a $k$-basis of $A$ containing 1 and working with "structure constants" for the $k$-algebra structure on $A$ relative to that basis.
Our task is to show that $I(k)$ is non-empty if and only if $I(K)$ is non-empty. Note that $I(K) = I_K(K)$. By the Nullstellensatz, since $k$ and $K$ are algebraically closed and $I$ is a $k$-scheme of finite type, this is equivalent to saying that $I$ is non-empty if and only if $I_K$ is non-empty. But this final equivalence is clear (as it expresses the non-vanishing of a coordinate ring, and non-vanishing of an algebra over a field is insensitive to extension of the ground field).
[The use of $I$ could be expressed entirely ring-theoretically, so its true power only becomes apparent in a wider context; e.g. one can use deformation theory to show that $I$ is actually $k$-smooth, so it yields that any absolutely semisimple algebra over a field becomes a matrix algebra over a finite etale extension of the field, a classical fact which is nonetheless understood in a more conceptual manner via the smoothness of $I$ and thereby adapts to a useful technique in the relative setting over rings and schemes.]
Proof of (ii): This will be all about "denominator-chasing" (the Nullstellensatz hiding in the way we will invoke (i) below).
Since $k = {\rm{Frac}}(B)$ and $A = R_k$ is a free $k$-module, by denominator-chasing with a choice of $k$-basis we can find a nonzero $b_0 \in B$ such that the subset $R[1/b_0] \subset R_k$ is spanned by a $B[1/b_0]$-linearly independent set, which is to say $R[1/b_0]$ is $B[1/b_0]$-free. We may replace $B$ with $B[1/b_0]$ and $R$ with $R[1/b_0]$ to reduce to the case that $R$ is $B$-free.
By (i), it now suffices to find a nonzero $b \in B$ and a $B[1/b]$-algebra $C$ such that ${\rm{Spec}}(C) \rightarrow {\rm{Spec}}(B[1/b])$ is surjective and $R_{C} := R \otimes_B C$ is a product of matrix algebras over $B'$. There is certainly a finite extension $k'/k$ such that $A_{k'}$ is a product of matrix algebras.
Let $B' \subset k'$ be a module-finite $B$-algebra such that $B'_k = k'$ (easy to build by denominator-chasing with a $k$-basis of $k'$). Since $B'_k$ is $k$-free, there is a nonzero $b_1 \in B$ such that $B'[1/b_1]$ is $B[1/b_1]$-free. By localizing at $b_1$ we may then assume $B'$ if $B$-free, so ${\rm{Spec}}(B') \rightarrow {\rm{Spec}}(B)$ is open. Hence, we may replace $(k, B, R)$ with $(k', B', R_{B'})$ to reduce to the case
that $A=R_k$ is a direct product of matrix algebras and $R$ is $B$-free.
Fix an isomorphism of $k$-algebras
$$k \otimes_B R= A \simeq \prod {\rm{Mat}}_{n_i}(k) =
k \otimes_B \prod {\rm{Mat}}_{n_i}(B).$$
Since $k = {\rm{Frac}}(B)$ and both $R$ and $\prod {\rm{Mat}}_{n_i}(B)$ are free as $B$-modules, by expressing the above $k$-algebra isomorphism
relative to $B$-bases of each we can denominator-chase to find a nonzero $b \in B$ such that the isomprphism arises from a $B[1/b]$-algebra isomorphism
$$R[1/b] \simeq \prod {\rm{Mat}}_{n_i}(B[1/b]).$$
