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.

I haven't thought about the case of a complex simple Lie group modulo its maximal compact, but the natural first case to check would be $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.

I should, perhaps, remark, that 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 $G=\mathrm{SO}(1,n)$.)

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 $G=\mathrm{SO}(1,n)$.)

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 two homogeneous copies of the hyperbolic disc that are minimal surfaces: The first, which is totally geodesic, consists of the 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 not totally geodesic, but is minimal (because it is a complex curve in $\mathbb{B}^2$), is the set of 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. It is not totally geodesic because it is not the set of $h$-positive lines lying in a $2$-dimensional plane in $\mathbb{C}^3$.

I haven't thought about the case of a complex simple Lie group modulo its maximal compact, but the natural first case to check would be $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.

I should, perhaps, remark, that 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 $S^n$, TAMS, 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 $G=\mathrm{SO}(1,n)$.)

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.

I haven't thought about the case of a complex simple Lie group modulo its maximal compact, but the natural first case to check would be $\mathrm{SL}(3,\mathbb{C})/\mathrm{SU}(3)$.

I should, perhaps, remark, that 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 $G=\mathrm{SO}(1,n)$.)