The answer is yes.
A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)
Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.
