Let me try another possible example with no nontrivial solution:

Let $\Sigma$ be a compact Riemann surface of genus $2$ with an antiholomorphic involution $\iota:\Sigma\to \Sigma$ whose fixed point locus consists of $3$ disjoint circles $C_i$ ($i=1,2,3$) whose complement consists of two disjoint connected regions, say $\Sigma_+$ and $\Sigma_-$. Equip $\Sigma$ with the unique conformal metric $g$ with Gauss curvature $K\equiv-1$. Let $M$ be the closure of $\Sigma_+$, so that $M$ is the union of $\Sigma_+$ and the circles $C_i$, which constitute the boundary of $M$. Then, because these are the fixed points of $\iota$ which is a $g$-isometry, each of the $C_i$ is a geodesic, i.e., $\kappa\equiv0$.

Now suppose that $f$ is defined and smooth on $M$ and satisfies $\Delta_gf = -2K f = 2f$ and $\partial_\nu f = \kappa f = 0$ on the boundary circles $C_i$. Then, extending the definition of $f$ to $\Sigma_-$ by setting $f(s) = f\bigl(\iota(s)\bigr)$ for $s\in \Sigma_-$, we get an $f$ defined on all of $\Sigma$ (and invariant under $\iota$), continuously differentiable along the curves $C_i$, and that satisfies $\Delta_g f = 2f$. (Here, I am using a reflection principle based on the fact that the normal derivative of $f$ along the boundary circles is zero, to get the necessary regularity of $f$ across the boundary.)

However, $\Sigma$ is compact and $\Delta_g$ has no positive eigenvalues, so $f$ must be identically zero.

(If your Laplacian has the opposite sign from mine, the same sort of argument will work to give a counterexample on the upper hemisphere of the unit sphere, since, in that case, $K\equiv+1$ and $\kappa\equiv0$ along the boundary.)

Added remark: I just realized that my example above is more complicated than it has to be: One can also find a compact Riemann surface of genus $2$ with an antiholomorphic involution that has only one fixed circle (think of making it by symmetrically connect-summing two $1$-holed tori along a fixed circle). Thus, one can construct such a counterexample on an $M$ with $K\equiv-1$ for which the (geodesic) boundary is connected.

Here is an example with no nontrivial solution:

Let $\Sigma$ be a compact Riemann surface of genus $2$ with an antiholomorphic involution $\iota:\Sigma\to \Sigma$ whose fixed point locus consists of $3$ disjoint circles $C_i$ ($i=1,2,3$) whose complement consists of two disjoint connected regions, say $\Sigma_+$ and $\Sigma_-$. Equip $\Sigma$ with the unique conformal metric $g$ with Gauss curvature $K\equiv-1$. Let $M$ be the closure of $\Sigma_+$, so that $M$ is the union of $\Sigma_+$ and the circles $C_i$, which constitute the boundary of $M$. Then, because these are the fixed points of $\iota$ which is a $g$-isometry, each of the $C_i$ is a geodesic, i.e., $\kappa\equiv0$.

Now suppose that $f$ is defined and smooth on $M$ and satisfies $\Delta_gf = -2K f = 2f$ and $\partial_\nu f = \kappa f = 0$ on the boundary circles $C_i$. Then, extending the definition of $f$ to $\Sigma_-$ by setting $f(s) = f\bigl(\iota(s)\bigr)$ for $s\in \Sigma_-$, one obtains an analytic $f$ defined on all of $\Sigma$ (and invariant under $\iota$), that satisfies $\Delta_g f = 2f$. (Here, I am using a reflection principle based on the fact that the normal derivative of $f$ along the boundary circles is zero, to get the necessary regularity of $f$ across the boundary. The point is that elliptic regularity coupled with the fact that both the metric and the boundary are real-analytic shows that $f$ is real-analytic up to and including the boundary circles, and then solving the analytic initial value problem along the boundary circles with the initial values that $f$ and $\partial_\nu f$ $(=0)$ are specified along the circles shows that $f$ extends analytically to a neighborhood of the circles in $\Sigma$ and must be $\iota$-symmetric there because the initial conditions are $\iota$-invariant.)

However, $\Sigma$ is compact and $\Delta_g$ has no positive eigenvalues, so $f$ must be identically zero.

(If your Laplacian has the opposite sign from mine, the same sort of argument will work to give a counterexample on the upper hemisphere of the unit sphere, since, in that case, $K\equiv+1$ and $\kappa\equiv0$ along the boundary.)

Added remark: I just realized that my example above is more complicated than it has to be: One can also find a compact Riemann surface of genus $2$ with an antiholomorphic involution that has only one fixed circle (think of making it by symmetrically connect-summing two $1$-holed tori along a fixed circle). Thus, one can construct such a counterexample on an $M$ with $K\equiv-1$ for which the (geodesic) boundary is connected.

Let me try another possible example with no nontrivial solution:

Let $\Sigma$ be a compact Riemann surface of genus $2$ with an antiholomorphic involution $\iota:\Sigma\to \Sigma$ whose fixed point locus consists of $3$ disjoint circles $C_i$ ($i=1,2,3$) whose complement consists of two disjoint connected regions, say $\Sigma_+$ and $\Sigma_-$. Equip $\Sigma$ with the unique conformal metric $g$ with Gauss curvature $K\equiv-1$. Let $M$ be the closure of $\Sigma_+$, so that $M$ is the union of $\Sigma_+$ and the circles $C_i$, which constitute the boundary of $M$. Then, because these are the fixed points of $\iota$ which is a $g$-isometry, each of the $C_i$ is a geodesic, i.e., $\kappa\equiv0$.

Now suppose that $f$ is defined and smooth on $M$ and satisfies $\Delta_gf = -2K f = 2f$ and $\partial_\nu f = \kappa f = 0$ on the boundary circles $C_i$. Then, extending the definition of $f$ to $\Sigma_-$ by setting $f(s) = f\bigl(\iota(s)\bigr)$ for $s\in \Sigma_-$, we get an $f$ defined on all of $\Sigma$ (and invariant under $\iota$), continuously differentiable along the curves $C_i$, and that satisfies $\Delta_g f = 2f$. (Here, I am using a reflection principle based on the fact that the normal derivative of $f$ along the boundary circles is zero, to get the necessary regularity of $f$ across the boundary.)

However, $\Sigma$ is compact and $\Delta_g$ has no positive eigenvalues, so $f$ must be identically zero.

(If your Laplacian has the opposite sign from mine, the same sort of argument will work to give a counterexample on the upper hemisphere of the unit sphere, since, in that case, $K\equiv+1$ and $\kappa\equiv0$ along the boundary.)

Let me try another possible example with no nontrivial solution:

Let $\Sigma$ be a compact Riemann surface of genus $2$ with an antiholomorphic involution $\iota:\Sigma\to \Sigma$ whose fixed point locus consists of $3$ disjoint circles $C_i$ ($i=1,2,3$) whose complement consists of two disjoint connected regions, say $\Sigma_+$ and $\Sigma_-$. Equip $\Sigma$ with the unique conformal metric $g$ with Gauss curvature $K\equiv-1$. Let $M$ be the closure of $\Sigma_+$, so that $M$ is the union of $\Sigma_+$ and the circles $C_i$, which constitute the boundary of $M$. Then, because these are the fixed points of $\iota$ which is a $g$-isometry, each of the $C_i$ is a geodesic, i.e., $\kappa\equiv0$.

Now suppose that $f$ is defined and smooth on $M$ and satisfies $\Delta_gf = -2K f = 2f$ and $\partial_\nu f = \kappa f = 0$ on the boundary circles $C_i$. Then, extending the definition of $f$ to $\Sigma_-$ by setting $f(s) = f\bigl(\iota(s)\bigr)$ for $s\in \Sigma_-$, we get an $f$ defined on all of $\Sigma$ (and invariant under $\iota$), continuously differentiable along the curves $C_i$, and that satisfies $\Delta_g f = 2f$. (Here, I am using a reflection principle based on the fact that the normal derivative of $f$ along the boundary circles is zero, to get the necessary regularity of $f$ across the boundary.)

However, $\Sigma$ is compact and $\Delta_g$ has no positive eigenvalues, so $f$ must be identically zero.

(If your Laplacian has the opposite sign from mine, the same sort of argument will work to give a counterexample on the upper hemisphere of the unit sphere, since, in that case, $K\equiv+1$ and $\kappa\equiv0$ along the boundary.)

Added remark: I just realized that my example above is more complicated than it has to be: One can also find a compact Riemann surface of genus $2$ with an antiholomorphic involution that has only one fixed circle (think of making it by symmetrically connect-summing two $1$-holed tori along a fixed circle). Thus, one can construct such a counterexample on an $M$ with $K\equiv-1$ for which the (geodesic) boundary is connected.