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.