Ok, so here it goes.

First, let us do the $M_2(\mathbb{C})$ case. Let $t\in(0,1)$, and define
$$
p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix}.
$$
Note that $p\wedge q=0$, since their ranges are two distinct lines through the origin.
Define a (faithful) state $\varphi$ by
$$
\varphi\left(\begin{bmatrix}a&b\\ c&d\end{bmatrix}\right)=\frac{2a+d}3
$$
(any convex combination $ra+sd$ with $r>s$ will do). Now
$$
\varphi\left(p+q\right)=\frac{2(1+t)+1-t}3=1+\frac{t}3>1.
$$
So that's the counterexample in $M_2(\mathbb{C})$

If now $M$ is a II$_1$ factor, we can use the same idea in the following way: let $p$ be any projection of trace 1/2. Then $p\sim(1-p)$ and there exists a partial isometry $v\in M$ with $v^*v=p$, $vv^*=1-p$. The four operators $p,v^*,v,1-p$ behave exactly as the matrix units $e_{11},e_{12},e_{21},e_{22}.$ So we define $q=tp+\sqrt{t-t^2}(v+v^*)+(1-t)(1-p)$, which is a projection; it is easy to check that $\tau(v)=0$, and that $\tau(q)=1/2$. Let $\varphi$ be the (faithful) state $\varphi(x)=2\tau(2px+(1-p)x)/3$. Then
$$
2p(p+q)+(1-p)(p+q)=2p+2pq+p+q-p-pq=2p+pq+q,
$$
and
$$
\varphi(p+q)=\frac23\,\tau(2p+pq+q)>\frac23\,\tau(2p+q)=\frac23\,\left(1+\frac12\right)=1.
$$
It remains to see that $p\wedge q=0$. Represent $M$ faithfully on a Hilbert space $H$. Suppose that $\xi\in pH\cap qH$. Then $\xi=p\xi=q\xi$. In particular, $(1-p)\xi=0$. Then $v^*\xi=0$, and so
$$
\xi=q\xi=tp\xi+\sqrt{t-t^2}v\xi.
$$
The last piece of information we need is that $v=(1-p)v$. Then $pv\xi=0$, and
$$
\xi=p\xi=pq\xi=tp\xi=t\xi.
$$
Since $t\ne1$, this forces $\xi=0$.

`II$_1$`

factor. – Martin Argerami Nov 11 '10 at 22:48