I thought about it, and the second answer I proposed earlier is not all that bad. I'm going to treat the points $A$, $B$, $C$, $Q$, as coordinate triples, so that I can add and subtract them as vectors. I'm also going to write $r_2$ for $R$, the radius of the second sphere, and $r_1$ for the radius of the first. Here goes:
$$
\newcommand{\a}{\mathbf a}
\newcommand{\b}{\mathbf b}
\newcommand{\c}{\mathbf c}
\newcommand{\u}{\mathbf u}
\newcommand{\v}{\mathbf v}
\newcommand{\w}{\mathbf w}
$$
\begin{align}
r_1 &= \|A - C\| \\
r_2 & = R\\
\a &= (A - Q)/r_2 \\
\b &= (B - Q)/r_2\\
\c &= (C - Q)/r_2.
\end{align}
In doing this, we've changed to a coordinate system in which the second sphere is at the origin, and has radius 1; we'll convert back at the end.
Let
\begin{align}
\u &= \frac{r_2}{r_1}(\a - \c) \\
\v &= \frac{r_2}{r_1}(\b - \c) \\
\v' &= \v - (\u \cdot \v) \u \\
\w & = \v' / \| \v' \|
\end{align}
Note that $\|u\| = \|v \| = \|w\| = 1$. And $\u \cdot \w = 0$.
Now the arc from $\u$ to $\w$ passes through $\v$ at some angle $T$. What angle?
\begin{align}
T &= \arccos ({\v \cdot \u})
\end{align}
Let
\begin{align}
\gamma(t) = \c + r_1\cos(t) \u + r_1\sin(t) \w
\end{align}
for $0 \le t \le T$.
Then $\gamma$ parameterizes the arc from $\a$ to $\b$ on the (translated and scaled) first sphere. We seek points where
\begin{align}
\| \gamma(t) \|^2 &= 1.
\end{align}
because those are the ones that lie on the second sphere (in the new coordinate system).
Writing this out, we have
\begin{align}
1 &= \| \gamma(t) \|^2 \\
&= \gamma(t) \cdot \gamma(t)\\
&= (\c + r_1\cos(t) \u + r_1\sin(t) \w) \cdot (\c + r_1\cos(t) \u + r_1\sin(t) \w) \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1\sin(t) (\w \cdot \c) + r_1^2\cos^2(t) (\u \cdot \u) + 2r_1^2\cos(t)\sin(t) (\u \cdot \w) + r_1^2\sin^2(t) (\w \cdot \w)
\end{align}
Fortunately, $\u$ and $\w$ are orthogonal, and each of $\u$ and $\w$ is a unit vector. So we have
\begin{align}
1 &= \| \gamma(t) \|^2 \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1\sin(t) (\w \cdot \c) + r_1^2\cos^2(t) + r_1^2\sin^2(t) \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1 \sin(t) (\w \cdot \c) + r_1^2 \\
\end{align}
Now we write $s = \tan(\frac{t}{2})$, which turns out to mean that $\sin t = \frac{2s}{1+s^2}$ and $\cos t = \frac{1-s^2}{1+s^2}$. So our equation becomes
\begin{align}
1
&= (\c\cdot \c) + 2r_1\frac{1-s^2}{1+s^2} (\u \cdot \c) + 2 r_1\frac{2s}{1+s^2} (\w \cdot \c) + r_1^2 \\
\end{align}
Multiplying through by $1+s^2$ gives
\begin{align}
1+s^2
&= (1+s^2)(\c\cdot \c) + 2r_1(1-s^2)(\u \cdot \c) + 4r_1s (\w \cdot \c) + r_1^2(1+s^2) \\
0
&= -1 -s^2 + (1+s^2)(\c\cdot \c) + 2r_1(1-s^2)(\u \cdot \c) + 4r_1s (\w \cdot \c) + r_1^2(1+s^2) \\
0 &= s^2 (\|c\|^2 - 1 - 2 r_1\u \cdot \c + r_1^2) + s 4r_1 (\w \cdot \c) + 2r_1(\u \cdot \c) + (\c \cdot \c) - 1 + r_1^2
\end{align}
which is a quadratic $As^2 + Bs + C = 0$, where the coefficients are
\begin{align}
A &= (\|c\|^2 - 1 - 2 r_1u \cdot c + r_1^2)\\
B &= 4r_1(w \cdot c)\\
C &= 2r_1(u \cdot c) + (c \cdot c) - 1 + r_1^2
\end{align}
(My apologies here for reusing the names $A, B, C$.) This quadratic can be solved with the quadratic formula to get solutions $s_1, s_2$. If both solutions have an imaginary part (which happens when $B^2 - 4AC < 0$), then the arc does not intersect the sphere. Otherwise, they are real, but may be identical. Let's see how to translate them back into an overall solution to the problem.
At this point, now that we've solved the quadratic, "A, B, C" go back to their original meanings as points in 3-space, OK?
From $s_i$ ($i = 1, 2$ henceforth), we can compute
$$
t_i = \arctan(2 s_i).
$$
If either value of $t_i$ is outside the interval $[0, T]$, we ignore it -- it does not correspond to a point of intersection of the arc and the sphere.
From this, we can compute $\cos t_i$ and $\sin t_i$, and thus compute
$$
\gamma_i = \c + r_1\cos t_i \u + r_1\sin t_i \w
$$
and then, going back to the original coordinate system, compute
$$
X_i = r_2\gamma_i + Q
$$
and the points $X_i$ are the intersections you seek.
