Lasse's answer expanded: Let $\psi$ be the map of the exterior of the unit disk onto the exterior of the Mandelbrot set, with Laurent series
$$
\psi(w) = w + \sum_{n=0}^\infty b_n w^{-n} =
w - \frac{1}{2} + \frac{1}{8} w^{-1} - \frac{1}{4} w^{-2} +
\frac{15}{128} w^{-3} + 0 w^{-4} -\frac{47}{1024} w^{-5} + \dots
$$
Then of course the boundary of the Mandelbrot set is the image of the unit circle under this map. However, this depends on the (not yet proved) local connectedness of that boundary. Here, for the coefficients $b_n$ there is no known closed form, but they can be computed recursively. Of course we put $w = e^{i\theta}$ and then this is a Fourier series.
Your $f_\lambda$ implements Newton's method for $p_\lambda$. As you said, the zeros $\{ 1,-1,\lambda \}$ of $p_\lambda$ are fixed points of $f_\lambda$, but even more, they are superattracting fixed points; that is, $f'_\lambda(1) = f'_\lambda(-1) = f'_\lambda(\lambda) = 0$.
If you choose a fixed $\lambda$ and iterate different points to see whether they are attracted to $1$, $-1$, or $\lambda$, coloring them accordingly, you will be plotting the Julia set of $f_\lambda$; this looks like the Newton fractals that you mention. Indeed, the first fractal in the Wikipedia page you linked to is the Julia set of the Newton method function associated to $z \mapsto z^3-1$.
Instead, what you are doing is plotting the results of iterating one particular value for many different parameters $\lambda$. This is not a dynamical procedure because you use a different $f_\lambda$ every time. This makes it very unlikely that your pictures are Julia sets. However, it can be shown that locally, near the critical value $f_{\lambda}(\lambda/3)$, the Julia set of $f_\lambda$ looks similar to this parametric picture, and this is why your pictures look like Newton fractals. For a proof of the similarity statement in the quadratic family, see Tan Lei's paper: Similarity between the Mandelbrot set and Julia sets
Comm. Math. Phys. Volume 134, Number 3 (1990), 587-617.
The original Mandelbrot set is constructed by iterating 0 with different parameters $c$ in the family $z \mapsto z^2+c$; a parametric construction just as above. The appearance of little Mandelbrots sets in your pictures signals a region of parameters $\lambda$ where an iterate of $f_\lambda$ has some (local) behavior that is conjugate to a quadratic polynomial. If you pick $\lambda$ in one of those small Mandelbrots and draw the Julia set of $f_\lambda$, you will find small regions that looks like the Julia set of that quadratic polynomial!
Added: Eremenko is right. Just because a picture is not generated by a dynamical process, it doesn't necessarily follow that it can't be a Julia set. A heuristic argument to support that conclusion is as follows: Even though Julia sets of rational functions have structure at all scales, this structure can be described by a sequence of finite combinatorial objects (for quadratic maps, we have kneading sequences and external angles). Once a picture contains something as complicated as a Mandelbrot set, it is possible to find infinitely many different such sequences of combinatorial data. It is in this sense that I say it is very unlikely that a parametric fractal like yours can be generated as a Julia set.
Best Answer
Your notation is unusual, and I am not sure whether I entirely understand it.
I shall take your question to mean: Can every point of the Mandelbrot $M$ set be connected to $\infty$ by a path that intersects the boundary $\partial M$ in only finitely many points?
This is connected to a very famous conjecture, namely that The Mandelbrot set is locally connected. (See The deep significance of the question of the Mandelbrot set's local connectedness?.)
Indeed, if the Mandelbrot set is locally connected, then - in particular - every point of the boundary of M is accessible from $\infty$. Note that "finitely many points" can be replaced by "one point". (Recall that the Mandelbrot set is full, and hence every interior component is simply connected.)
If the Mandelbrot set is not locally connected, then it follows e.g. from the theory of "fibers", as formulated by Dierk Schleicher, that there exists a point of the boundary that is not accessible from the complement, and hence the answer to your question would be negative.
(Caution. This statement is not true for general compact sets: a compact and full connected set can have every point accessible, but not be locally connected. The statement above uses the specific structure we know about the Mandelbrot set.)
If you start asking for curves with specific geometric properties, the question will become more subtle. For example, if you ask for smooth curves, the answer is 'no' in general (as the Mandelbrot set spirals at many points).