$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$**Remark 1:**
>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 $s\in I_P:=(-\la_{\min},\infty)$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. The function $g$ is nonnegative, nonincreasing, and $g(\infty-)=0$. 
Moreover, the equation 
\begin{equation*}
	g(s)=1 \tag{0}\label{0}
\end{equation*}
$s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots.  

Let now
\begin{equation*}
	f(z):=\frac12\,z^T Pz+q^T z+r
\end{equation*}
for $z\in\R^n$. 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=-(P+tI)^{-1}q, \tag{4}\label{4}
\end{equation*}
whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the only (and hence the largest) root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$. 

*Case 2*: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if  and only if $|x|=1$. 
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. 

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, in both cases, (i) equation \eqref{0} has at most one root and (ii) if it has a root, then this root is $t=|a|=|Px+q|=\max(0,t)$, for $x$ as above. 
 
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These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.