This is not true. E.g. the trace condition is satisfied when $n=2$ and
$$
K=\pmatrix{2&1\\ 1&1},\ u=\pmatrix{1\\ 1},\ v=\pmatrix{0\\ 1}.
$$
In general, as $(K+vv^T)^{-1}=K^{-1}-\frac{K^{-1}vv^TK^{-1}}{1+v^TK^{-1}v}$. If you put $P=K^{-1}$, using the tracial property, the trace condition can be rewritten as
$$
0=-\frac{v^TPv}{1+v^TPv}
+u^TPu-\frac{(u^TPv)^2}{1+v^TPv}
$$
or equivalently,
$$
(u^TPu)(1+v^TPv)-v^TPv-(u^TPv)^2=0.\tag{1}
$$
When $m_u\le m_v$, you may partition $P$ as
$$
\pmatrix{\ast&\ast&\ast\\ \ast&B&Z\\ \ast&Z^T&C}
$$
where $B$ is $(m_v-m_u)\times(m_v-m_u)$ and $C$ is $(n-m_v)\times(n-m_v)$. Let the sum of all entries in $B$ be $b$ and define $c$ and $z$ analogously for $C$ and $Z$. Condition $(1)$ can then be rewritten as
\begin{align}
&(b+c+2z)(1+c)-c-(c+z)^2=0\\
\Leftrightarrow\ &b(c+1)=z(z-2).\tag{2}
\end{align}
In the counterexample above, we have $P=K^{-1}=\pmatrix{1&-1\\ -1&2}$, so that $b=1,c=2$ and $z=-1$.
When $m_u>m_v$, you can partition $P$ analogously, but this time, $B$ is $(m_u-m_v)\times(m_u-m_v)$ and $C$ is $(n-m_u)\times(n-m_u)$. The trace condition in this case becomes
\begin{align}
&c(1+b+c+2z)-(b+c+2z)-(c+z)^2=0\\
\Leftrightarrow\ &b(c-1)=z(z+2).\tag{3}
\end{align}
So, even when $u\ne v$, it can happen that $(2)$ or $(3)$ hold when the values of $b,c,z$ are appropriate.
