$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Let 
\begin{equation}
	g(s):=q^T(P+sI)^{-2}q
\end{equation}
for real $s$ such that $P+sI$ is invertible, that is, for all real $s$ greater than $-\la_{\min}$, where $\la_{\min}$ is the smallest eigenvalue $\la_{\min}$ of the symmetric positive-definite matrix $P$. The function $g$ is nonnegative, nonincreasing, and $g(\infty-)=0$. 
Let 
\begin{equation}
	f(x):=\frac12\,x^T Px+q^T x+r
\end{equation}
for $x\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have 
\begin{equation}
	0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1}
\end{equation}
where 
\begin{equation}
	a:=\nabla f(x)=Px+q \tag{2}\label{2}
\end{equation}
and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. 
So, by \eqref{1},
\begin{equation}
 -a^T x=|a|.  \tag{3}\label{3}	
\end{equation}
So, for 
$$t:=|a|,$$
we have one of the following two cases: 

*Case 1:* $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and  
\begin{equation}
	x=x_t, \quad\text{where } x_s:=-(P+sI)^{-1}q, 
\end{equation}
whence $g(t)=x^T x=1$ and hence 
\begin{equation}
	f(x)=h(t), 
\end{equation}
where 
\begin{equation}	
\begin{aligned}
	h(s)&:=f(x_s)=\frac12\,q^T (P+sI)^{-1} P(P+sI)^{-1}q-q^T (P+sI)^{-1}q+r \\ 
	&:=-\frac12\,q^T (P+sI)^{-1}q-\frac s2\,q^T (P+sI)^{-2}q+r \\ 
	&:=-\frac12\,q^T (P+sI)^{-1}q-\frac s2\,g(s)+r. 
\end{aligned}
\end{equation}
Recall that the continuous function $g$ is nonincreasing. So, the set $S:=\{s\in\R\colon g(s)=g(t)\}=\{s\in\R\colon g(s)=1\}$ is of the form $[t_1,t_2]\cap(-\la_{\min},\infty)$ for some real $t_1,t_2$ such that $t_2\ge t$. So, for all $s$ in the interior of the set $S$ we have 
\begin{equation}	
\begin{aligned}
	h'(s)&=\frac12\,q^T (P+sI)^{-2}q-\frac 12\,g(t)=0. 
\end{aligned}
\end{equation}
So, $f(x_s)=h(s)=h(t)=f(x)$ for all $s\in S$. Therefore and because $x=x_t$ is the unique minimizer of $f$ on $B$, we conclude that $x_s=x_t=x$ for all $s\in S$. In particular, $x_{t_2}=x$. 
So, in Case 2, the minimizer  of $f$ on $B$ is 
\begin{equation}
	x=x_{t_2}=-(P+t_2I)^{-1}q, 
\end{equation}
and $t_2>0$ is the largest real root $s$ of the equation 
\begin{equation}
	g(s)[=q^T(P+sI)^{-2}q]=1,
\end{equation}
as desired. 

*Case 2*: $t=0$, that is, $x=-P^{-1}q$. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(0)\le1$. However, the equation $g(s)=1$ may not have any real root $s$ at all (e.g. when $q=0$ or, more generally, when $q$ is a short enough nonzero vector orthogonal to the eigenspace of $P$ corresponding to $\la_{\min}$). So, in this case your claim about the minimizer will not hold in general. 
 
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These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and uisng the spherical symmetry of $B$.