If you want your subalgebra to be unital, it is not true. For example, let $A=\mathbb C\oplus M_3(\mathbb C)$. Then the only proper unital subalgebras are $\mathbb C\oplus \mathbb C$ and---up to unitary equivalence---$\mathbb C\oplus M_2(\mathbb C)\oplus\mathbb C$ (note that $0\oplus M_2(\mathbb C)\oplus 0$ is a subalgebra, but not a unital one).
If you allow the subalgebra to have a different unit than $A$, the answer is always yes. Note that we can characterize $M_2(\mathbb C)$ as the span of $e_{11}, e_{12}, e_{21}, e_{22}$, where $e_{11}$ and $e_{22}$ are projections with $$e_{11}e_{22}=0,\ \ \ \ \ e_{21}=e_{12}^*,\ \ \ \ \ e_{12}e_{21}=e_{11},\ \ \ \ \ e_{21}e_{12}=e_{22}$$ (these would be the canonical matrix units).
Argument. Because $A\subset B(H)$ is non-abelian, there exists a non-central projection $p\in A$. Because $p$ is non central, it cannot be equal to its central carrier $c(p)$. So $q=c(p)-p$ is a nonzero projection, and $pq=0$. Since $q\leq c(p)$, we have $c(q)\leq c(p)$.
So the subspaces $MpH$ and $MqH$ are not orthogonal (these are precisely the ranges of $c(p)$ and $c(q)$).
Claim. there exists $x\in A$ with $pxq\ne0$. Indeed, if $pxq=0$ for all $x\in A$, then $pMqH=0$. So $pH\perp MqH$. For any $y,z\in A$, $\xi,\eta\in H$, $$\langle yp\xi,zq\eta\rangle=\langle p\xi,y^*zq\eta\rangle=0.$$ Thus $MpH\perp MqH$, a contradiction.
So, let $x\in A$ such that $pxq\ne0$. Let $pxq=vh$ be the polar decomposition. Let
$$
e_{11}=v^*v,\ \ e_{12}=v^*,\ \ e_{21}=v,\ \ e_{22}=vv^*.
$$
Since $v$ is a partial isometry, $e_{11}$ and $e_{22}$ are projections, and $e_{12}e_{21}=v^*v=e_{11}$, $e_{21}e_{12}=vv^*=e_{22}$. It remains to see that $e_{11}e_{22}=0$. This amounts to see that $v^2=0$. From the polar decomposition, $vv^*$ is the range projection of $pxq$, so $pv=v$. Similarly, $v^*v$ is the range projection of $qx^*p$, so $qv^*=v^*$. Then $v=pvq$, and so
$$
v^2=pvqpvq=0.
$$
We now have that
$$
\text{span}\,\{v^*v,v^*,v,vv^*\}\subset A
$$
is isomorphic to $M_2(\mathbb C)$.
