When $k$ is odd, there does exist such a family of immersions satisfying Paul's requirements. (When $k$ is even, Vitali Kapovich has shown, using a clever topological argument, that it's not possible to have such a family of degenerating immersions. Please see his answer for the details.)
Set $k=2m+1$, and consider the (complex) $1$-parameter family of maps $u_t:\mathbb{C}\to\mathbb{R}^3$ given by $$u_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr).$$ These smooth maps converge smoothly to $u_0$ as $t\to0$, and $u_t$ induces the metric $$ds_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2.$$ Thus, $u_t$ is an immersion for $t\not=0$, while $u_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$.
The family $u_t$ was constructed using the Weierstrass formula for minimal immersions, so the image $u_t$ u_t({\mathbb{C}})\subset \mathbb{R}^3$is an immersed minimal surface, and, as a result, the Gauss curvature is everywhere non-positive. In fact, for$t\not=0$, the Gauss curvature only vanishes at$z=0$, and then only when$m>1$. In particular, all of these degenerating immersions have curvature bounded above. 1 When$k$is odd, there does exist such a family of immersions satisfying Paul's requirements. (When$k$is even, Vitali Kapovich has shown, using a clever topological argument, that it's not possible to have such a family of degenerating immersions. Please see his answer for the details.) Set$k=2m+1$, and consider the (complex)$1$-parameter family of maps$u_t:\mathbb{C}\to\mathbb{R}^3$given by $$u_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr).$$ These smooth maps converge smoothly to$u_0$as$t\to0$, and$u_t$induces the metric $$ds_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2.$$ Thus,$u_t$is an immersion for$t\not=0$, while$u_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$. The family$u_t$was constructed using the Weierstrass formula for minimal immersions, so the image$u_t$is an immersed minimal surface, and, as a result, the Gauss curvature is everywhere non-positive. In fact, for$t\not=0$, the Gauss curvature only vanishes at$z=0$, and then only when$m>1\$. In particular, all of these degenerating immersions have curvature bounded above.