The characteristic polynomial of $\left(\begin{array}{rr} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right)$ is, as you note, $t^2 -2(\cos\theta)t + 1$.
The discriminant of this polynomial is $4(\cos^2\theta) - 4 = 4(\cos^2\theta - 1).$ Therefore, the discriminant is always nonpositive, and is equal to zero if and only if $\cos^2\theta = 1$, if and only if $\theta = n\pi$ for some $n\in\mathbb{Z}$.
In particular, if $\theta\neq n\pi$ then the characteristic polynomial is irreducible over $\mathbb{Q}$, so the minimal polynomial agrees with the characteristic polynomial, and so the rational canonical form is just the companion matrix of the characteristic polynomial (just as you wrote): $\left(\begin{array}{rr} 0 & -1\\ 1 & 2\cos\theta \end{array}\right).$
If $\theta=n\pi$, then $A$ is one of the following matrices: $\begin{align*} A &= \left(\begin{array}{rr} 1 & 0\\ 0 & 1 \end{array}\right) &\text{if }n\text{ is even,}\\ A &=\left(\begin{array}{rr} -1 & 0\\ 0 & -1 \end{array}\right) &\text{if }n\text{ is odd.} \end{align*}$
Both of these matrices are already in rational canonical form.
(Your error: the minimal polynomial is indeed $t-\cos\theta$ in this situation; since the characteristic polynomial is then the square of the minimal polynomial, the rational canonical form will have two blocks, each associated to a degree 1 polynomial; not just a single block. Remember that the exponent of the irreducible factor on the minimal polynomial gives you the size of the largest block, in this case $1\times 1$; but the sum of the sizes of the blocks has to add up to the exponent of the irreducible factor in the characteristic polynomial, in this case $2$. So you need two $1\times 1$ blocks here. )