Yes. Since $A = A^2 + A^T A = A^2 + A A^T$, it follows that $A$ is normal, hence by the spectral theorem has an orthonormal basis of eigenvectors $v_1, ... v_n$. Moreover, the given condition says precisely that the corresponding eigenvalues $\lambda_i$ satisfy $A v_i + A^T v_i = \lambda v_i + \overline{\lambda} v_i = v_i$, hence all have real part exactly $\frac{1}{2}$. Since $A$ is real, the set of its eigenvalues is closed under complex conjugation, so every eigenvalue $\frac{1}{2} + iy$ is paired with its complex conjugate $\frac{1}{2} - iy$ (and the eigenvalue $\frac{1}{2}$ can occur with any multiplicity). The product of these is positive, so $\det A > 0$.
Here is an alternate argument which does not rely on the spectral theorem. The condition gives $\langle v, v \rangle = \langle (A + A^T) v, v \rangle = \langle Av, v \rangle + \langle A^T v, v \rangle = 2 \langle Av, v \rangle$
for $v \in \mathbb{R}^n$. It follows that any such matrix $A$ is invertible, since $Av = 0$ implies $\langle v, v \rangle = 0$. On the other hand, the set of all matrices $A$ satisfying this condition is an affine subspace - in particular, it is path-connected. A path-connected set of invertible matrices must lie in a path component of $\text{GL}_n(\mathbb{R})$, hence the determinant must be the same sign for every element of the set. And $\frac{1}{2} I$ is in this set.
Note that the set of matrices satisfying this condition is precisely a translate of the set of skew-symmetric matrices by $\frac{1}{2} I$.