Do there exist non-totally geodesic isometric minimal immersions $\mathbb{H}^2\rightarrow G/K.$ Suppose $G$ is a non-compact, semi-simple Lie group, of rank at least two, with maximal compact subgroup $K$ and $G/K$ the corresponding Riemannian symmetric space.  Let $\mathbb{H}^2(-c^2)$ be the 2-dimensional real hyperbolic space of sectional curvature $-c^2.$  Suppose,
\begin{align}
f:\mathbb{H}^2(-c^2)\rightarrow G/K,
\end{align}
is an isometric, minimal (i.e. vanishing mean curvature) immersion.  Is $f$ necessarily totally geodesic?  I'm primarily concerned in the case that G is a complex simple Lie group, but perhaps the statement is true in this broader context.  Thanks for any help!  
 A: Below, I have added an answer to your question about the case of a complex semi-simple Lie group.  It turns out that the answer is 'no' even in this case.
The answer to your general question is 'no'.  A simple example is to take $G = \mathrm{SU}(1,2)$ with its maximal compact $K = \mathrm{U}(2)$.  You can think of the homogeneous space $G/K$ as the unit ball $\mathbb{B}^2\subset \mathbb{C}^2$, but you can also think of it as the space of complex lines $L$ in $\mathbb{C}^3$ on which the Hermitian form $h =|z^0|^2-|z^1|^2-|z^2|^2$ is positive definite.  
When endowed with its canonical $G$-invariant Kähler metric, $\mathbb{B}^2=G/K$ contains three mutually noncongruent, homogeneous copies of the hyperbolic disc that are minimal surfaces:  The first, which is totally geodesic, consists of the $h$-positive complex lines $L$ that are complexifications of real lines in $\mathbb{R}^3$, i.e., those lines for which $L = \bar L$.  The second, which is also totally geodesic, consists of the $h$-positive complex lines $L$ that lie in a fixed complex $2$-plane $P\subset\mathbb{C}^3$ of $h$-hermitian type $(1,1)$.  The third, which is not totally geodesic but is minimal (because it is a complex curve in $\mathbb{B}^2$), is the set of $h$-positive lines $L$ that are null with respect to the complex inner product $q = (z^0)^2-(z^1)^2-(z^2)^2$, i.e., that satisfy $L\cdot L = 0$ (while $L\cdot \bar L > 0$).  This curve is an orbit of the subgroup $\mathrm{SO}(1,2)$ and hence is $\mathrm{SO}(1,2)/\mathrm{SO}(2)$, i.e., the hyperbolic plane. 
Here is where I have modified my original answer:
As for the case of a complex simple Lie group modulo its maximal compact, the natural first case to check would be $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$ (since the answer is 'yes' for the lowest possible case, $\mathrm{SL}(2,\mathbb{C})/\mathrm{SU}(2)\simeq H^3$).  However, right away, the above example shows that the answer is 'no' for $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.  The reason is that the noncompact real form $\mathrm{SU}(1,2)\subset \mathrm{SL}(3,\mathbb{C})$ acts on the quotient $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$ with orbit $\mathbb{B}^2 = \mathrm{SU}(1,2)/\mathrm{U}(2)$ (since $\mathrm{SU}(1,2)\cap\mathrm{SU}(3)\simeq \mathrm{U}(2)$), and this orbit is embedded as a totally geodesic submanifold of $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.  Since we already constructed an isometric, minimal-but-not-totally-geodesic immersion of the hyperbolic disc into $\mathbb{B}^2$ (see above), it follows that this yields an isometric, minimal-but-not-totally-geodesic immersion of the hyperbolic disc into $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.  Thus, the answer is 'no' even in the special case you care about.
Remark: Meanwhile, the answer to your question is 'yes' for $G = \mathrm{SO}(1,n)$.  This is in an old article of mine: Minimal surfaces of constant curvature in $S^n$, Trans. Amer. Math. Soc. 20 (1985), 259–271.  (Don't be fooled by the title; at the end of the article, I treat the case of hyperbolic $n$-space $H^n$ as well, which is the case $G=\mathrm{SO}(1,n)$.)
