One can even define a *holomorphic* (i.e. integrable almost complex) structure on this Grassmanian. To give the definition it is easier to consider instead $Gr_2^{-}(E^{n+2}_{2})$ which is obviously the same object.

**Definition.** We will identify the Grassmanian of two-planes with a part of the quadric in $\mathbb CP^{n+1}$. Namely, let $\mathbb C^{n+2}$ be the complexification of $\mathbb R^{n+2}$: $\mathbb C^{n+2}=\mathbb R^{n+2}\oplus i \mathbb R^{n+2}$. Let $x_1,...,x_{n+2}$ be the coordinates in $\mathbb R^{n+2}$, so that the quadratic form corresponding to the metric is $x_1^2+...+x_{n}^2-x_{n+1}^2-x_{n+2}^2$. Let $z_1,...,z_n$ be the complexified coordinates, and let $Q(z)=z_1^2+...+z_{n}^2-z_{n+1}^2-z_{n+2}^2$ be the complexified quadratic form.

Now, consider the quadric cone $Q(z)=0$, in $\mathbb C^{n+2}$ ($z=x+iy$, $x,y\in \mathbb R^{n+2}$).  
Notice that $Gr_2^{-}(E^{n+2}_{2})$ is naturally is an  open subset of the projectivisation of the cone $Q(z)=0$. Indeed, for any pair of orthogonal unite vectors  $x\perp y$ in $\mathbb R^2\subset \mathbb R^{n+2}$ we can associate a complex vector $z=x+iy$, $z\in \mathbb C^{n+2}$, satisfying $Q(z)=0$. The line $\lambda z\subset \mathbb C^{n+2}$  is independent on the choice of orthogonal unite vectors $x,y$ in $\mathbb R^2$.

It is easy to see that the complex structure constructed this way is invariant with respect to the action of isometries of $\mathbb R^{n+2}$ on $Gr_2^{-}(E^{n+2}_{2})$.

**PS** *holomorphic structure in local coordinates*.

In order to present a bit more explicitly the above complex structure on the Grassmanian we can proceed in two ways. First is just to take some meromorphic functions on the quadric and see what kind of function we get on the Grassmanian. Simplest meromorphic function would be $\frac{z_k}{z_l}=\frac{x_k+iy_k}{x_l+iy_l}$. Now if you take an isotropic two-plane spanned by two unit ortogonal vectors $x$ and $y$, the value of the above function on the plane is precisely    $\frac{x_k+iy_k}{x_l+iy_l}$ (where $x_k, x_l$ are the corresponding coordinates of the vector $x$, and $y_k, y_l$ are $k$th and $l$th coordinates of $y$). So, this tells you how to construct lots of meromorphic functions.

**Second way** would be to define explicitly the operator $J$ on the tangent space to a point of the Grassmanian. Recall that the tangent space to a Grassmanian of $k$-planes $V\subset \mathbb R^{n+2}$ at point $V$ is $Hom(V,\mathbb R^{n+2}/V)$. In our case $dim V=2$ and $V^{\perp}\cap V=0$, so the tangent space is can be seen as $T_V=Hom(V,V^{\perp})$.
We want to find a natural $J$ on $T_V$. The best thing to do is to find $J$ that is invariant under the action of the stabiliser of (oriented) $V$ in the orthogonal group. This stabiliser is $SO(2)\times SO(n)$. Denote by $I$ the operator $V\to V$ that rotates  $V$ by $90^0$. Now we can define $J$ as follows: for any $A\in Hom(V,V^{\perp})$, $J(A)=AI$. Obviously $J^2=-Id$, and $J$ commutes with  $SO(2)\times SO(n)$. Finally note that $J$ and $-J$ are the only two operators that satisfy these properties.


**PPS. The metric** The space $Gr_2^{-}(E^{n+2}_{2})$ is a *symmetric space*, it has a metric such that for each point $p$ there is an isometry fixing $p$ and inducing the map $-Id$ on $T_p$. To define the metric we just use the fact that $V$ and $V^{\perp}$ are Euclidean spaces and there is a unique up to constant quadratic form on $Hom(V,V^{\perp})$ invariant under $SO(2)\times SO(n)$. This defines you a metric on $T_V$. The involution fixing $V$ and preserving the metric is given by $V\oplus V^{\perp}\to -V\oplus V^{\perp}$ i.e. it is $-Id$ on $V$ and $Id$ on $V^{\perp}$.
 One place to read a bit about symmetric spaces (and plenty of other interesting things) is section 2.1 in the Donaldson's *Lectures on Lie groups and geometry*: http://www2.imperial.ac.uk/~skdona/LIEGROUPSCONSOL.PDF