So $R$ is actually $\mathbb{F}_p[X]/(\Phi_m(X))$.
Assume that $R$ is a product of fields. In particular, $R$ is reduced (that is the problem for general $m$); in other words, $\Phi_m$ mod $p$ is separable. But if $p|m$, using complex roots, we can see that $\Phi_m(X)=\Phi_{m/p}(X^p)$ so is not separable mod $p$ (it’s equal to $\Phi_{m/p}(X)^p)$.
Now, assume that $p$ does not divide $m$. Then $\Phi_m$ divides $X^m-1$ which is separable mod $p$, so $\Phi_m$ is separable mod $p$, and coprime mod $p$ to any $\Phi_d$ with $d|m$ and $d < m$.
In particular, if $F$ is a field of characteristics $p$ and $\omega \in F$ is a root of $\Phi_m$, then $\omega$ has multiplicative order $m$ – and conversely.
Let $\Phi_m=P_1\ldots P_l$ be the decomposition in products of irreducible polynomials mod $p$, with the $P_i$ being monic, irreducible mod $p$ and pairwise distinct. Then $R$ is the product of the $\mathbb{F}_p[X]/(P_i)$ (by the CRT) which are actually fields generated by an element which is a root of $\Phi_m$, so is an element of multiplicative order $m$.
To conclude, we show that if $F$ is a field of degree $d$ over $\mathbb{F}_p$ generated by an element $\omega$ of multiplicative order $m$, then $d$ is the multiplicative order $u$ of $p$ mod $m$.
This shows that all the $P_i$ have the same degree.
Indeed, $1=\omega^{|F^{\times}|}=\omega^{p^d-1}$ so that $m|p^d-1$ and $u|d$. But $F’=\{z \in F,\,z^{p^u}=z\}$ is a subfield of $F$ containing $\mathbb{F}_p$ and $\omega$ so is $F$. But $F’$ is the set of roots of the polynomial $X^{p^u}-X$ so $|F|=|F’| \leq p^u$ so $d \leq u$ and $d=u$.
There is a calculation-free proof that "the smallest field over which $g$ could split" is indeed the splitting field. The key lemma is that a finite field has a unique extension of cardinality $n$ for any $n$.
Indeed, let $F$ be a finite field of cardinality $q$ and let $K$ be a degree $n$ extension of $F$. The cardinality of $K^\times$ is $q^n - 1$, and so every element $K^\times$ satisfies the polynomial $x^{q^n} - x$. This polynomial is separable (the formal derivative is $-1$), so it has $q^n$ distinct roots in $K$ -- but the cardinality of $K$ is $q^n$, so its roots are exactly the elements of $K$. That means that this polynomial splits in $K$ and over no subfield of $K$, and therefore $K$ is the splitting field of this polynomial, which makes it unique up to isomorphism.
Best Answer
Note that $Q_r$ has no multiple roots, and its degree is $\phi(r)$. Let $\beta$ be one of its roots so that $\beta^r=1$, and $\beta^s=1$ if and only if $r$ divides $s$. If $k$ is the least positive integer such that $q^k\equiv 1\pmod r$, then $\beta^{q^k-1}=\left(\beta^r\right)^{(q^k-1)/r}=1$, but $\beta^{q^j-1}\ne1$ for all $j<k$. It follows that $k$ is the least positive integer such that $\beta\in\mathbb{F}_{q^k}$. The minimal polynomial of $\beta$ thus has degree $k$. Since $\beta$ is arbitrary, every irreducible factor of $f$ has degree $k$, and $f$ has $\phi(r)/k$ factors ($k$ is the order of $q$ in the multiplicative group modulo $r$, and $k$ divides the order of the group).