$\newcommand\R{\mathbb R}\newcommand\tz{\tilde z}\newcommand{\de}{\delta}$Yes, the minimizer $x_{A,b}$ of $\frac12 x^TAx + b^Tx$ subject to $Cx\le d$ is
continuous with respect to $A$ and $b$ -- provided that the set $X:=\{x\in\R^n\colon Cx\le d\}$ is nonempty (otherwise, such a minimizer does not exist).
Indeed, let $M^+$ denote the set of all (symmetric) positive definite $n\times n$ real matrices. In what follows, $A,A_k$ are in $M^+$, and $b,b_k$ are in $\R^n=\R^{n\times1}$, where $k$ is any natural number.
Using the substitution $x=f_{A,b}(z):=A^{-1/2}z-A^{-1}b$, we rewrite the problem as
\begin{equation*}
\text{minimize } z^T z \quad \text{over } z\in Z_{A,b},
\end{equation*}
where
\begin{equation*}
Z_{A,b}:=\{z\in\R^n\colon Cf_{A,b}(z)\le d\}=f_{A,b}^{-1}(X).
\end{equation*}
So, $Z_{A,b}$ is a non-empty closed subset of $\R^n$ and hence there is a unique minimizer, say $z_{A,b}$, of $z^T z$ over $z\in Z_{A,b}$. So, $x_{A,b}:=f_{A,b}(z_{A,b})=A^{-1/2}z_{A,b}-A^{-1}b$ is the unique minimizer of $\frac12 x^TAx + b^Tx$ subject to $Cx\le d$.
So, it remains to show that $z_{A,b}$ is continuous in $A,b$.
To do this, suppose that $A_k\to A$ and $b_k\to b$ (as $k\to\infty$). Then
\begin{equation*}
z_k:=f_{A_k,b_k}^{-1}(f_{A,b}(z_{A,b}))\in Z_{A_k,b_k} \tag{0a}\label{0a}
\end{equation*}
and
\begin{equation*}
z_k\to z_{A,b}, \tag{0b}\label{0b}
\end{equation*}
so that
\begin{equation*}
z_k^T z_k\to z_{A,b}^T z_{A,b}=m_{A,b}:=\min\{z^T z\colon z\in Z_{A,b}\}
\end{equation*}
and hence
\begin{equation*}
z_{A_k,b_k}^T z_{A_k,b_k}=m_{A_k,b_k}\le m_{A,b}+o(1); \tag{1}\label{1}
\end{equation*}
in particular, the sequence $(z_{A_k,b_k})$ is bounded.
Therefore and because $A_k\to A$ and $b_k\to b$, we see that for
$$\tz_k:=f_{A,b}^{-1}(f_{A_k,b_k}(z_{A_k,b_k}))$$
we have $\tz_k-z_{A_k,b_k}\to0$. Therefore and because $(z_{A_k,b_k})$ is bounded, we have $\tz_k^T \tz_k-z_{A_k,b_k}^T z_{A_k,b_k}\to0$.
Also, $\tz_k\in Z_{A,b}$. So,
\begin{equation*}
m_{A,b}\le \tz_k^T \tz_k=z_{A_k,b_k}^T z_{A_k,b_k}+o(1)=m_{A_k,b_k}+o(1). \tag{2}\label{2}
\end{equation*}
By \eqref{1} and \eqref{2},
\begin{equation*}
m_{A_k,b_k}\to m_{A,b}. \tag{3}\label{3}
\end{equation*}
To obtain a contradiction, suppose now that $z_{A_k,b_k}\not\to z_{A,b}$. Then, passing to a subsequence, without loss of generality (wlog) assume that $|z_{A_k,b_k}-z_{A,b}|\ge2\de$ for some $\de>0$ and all $k$, where $|\cdot|$ is the Euclidean norm. So, by \eqref{0b}, wlog
\begin{equation*}
|z_{A_k,b_k}-z_k|\ge\de \tag{4}\label{4}
\end{equation*}
for all $k$. Because (i) the set $Z_{A_k,b_k}$ is convex and (ii) the minimizer $z_{A_k,b_k}$ of $z^T z$ over $z\in Z_{A_k,b_k}$ is in $Z_{A_k,b_k}$ and (iii) $z_k\in Z_{A_k,b_k}$ by \eqref{0a}, we see that $w_k:=\frac12\,(z_{A_k,b_k}+z_k)\in Z_{A_k,b_k}$. Using now (i) the parallelogram identity and (ii) the definitions of $z_{A_k,b_k}$ and $m_{A,b}$ and (iii) \eqref{0b} and \eqref{4}, we get
\begin{equation*}
\begin{aligned}
4|w_k|^2&=2|z_{A_k,b_k}|^2+2|z_k|^2-|z_{A_k,b_k}-z_k|^2 \\
&\le 2m_{A_k,b_k}+2(m_{A,b}+o(1))-\de^2 \\
&\le 4m_{A_k,b_k}+o(1)-\de^2,
\end{aligned}
\end{equation*}
so that for all large enough $k$ we have $w_k^T w_k=|w_k|^2<m_{A_k,b_k}$,
which contradicts the condition $w_k\in Z_{A_k,b_k}$, in view of the definition of $m_{A_k,b_k}$.
Thus, $z_{A_k,b_k}\to z_{A,b}$, which proves that $z_{A,b}$ is continuous in $A,b$. $\quad\Box$