I think the answer is ``No'' for $p$ odd.
Stronger yet, there exists no $\lambda$-ring structure on $\mathbb{Z}_{p}$
under which the canonical quotient map $W\left( \mathbb{F}_{p}\right)
\rightarrow\mathbb{Z}_{p}$ becomes a $\lambda$-ring morphism. However, due to
the manifold occasions for mistakes and misunderstandings in my argument,
please check it thoroughly.

Let $p$ be an odd prime. Let $\Lambda$ be the Hopf algebra of symmetric
functions over $\mathbb{Z}$ (in the indeterminates $x_{1},x_{2},x_{3},\ldots
$). It is known that $\Lambda$ carries further structures, such as plethysm,
inner multiplication and inner comultiplication. I am using the material of
Michiel Hazewinkel's *Witt vectors, part 1* here (although not always his
notations: my $\Lambda$ is his $\mathbf{Symm}$, and my $x_{1},x_{2}
,x_{3},\ldots$ are his $\xi_{1},\xi_{2},\xi_{3},\ldots$).

Fix a commutative ring $\mathbf{k}$. The $\lambda$-ring of big Witt vectors
$W\left( \mathbf{k}\right) $ is defined as the set $\operatorname*{Ring}
\left( \Lambda,\mathbf{k}\right) $ of all ring homomorphisms from
$\Lambda\rightarrow\mathbf{k}$, endowed with an addition that comes from the
comultiplication of $\Lambda$, an additive inverse that comes from the
antipode of $\Lambda$, a multiplication that comes from the inner
comultiplication of $\Lambda$, and a $\lambda$-ring structure that comes from
the plethysm of $\Lambda$.

We let $p_{1},p_{2},p_{3},\ldots$ be the power sum symmetric functions in
$\Lambda$ (that is, $p_{n}=\sum_{i\geq1}x_{i}^{n}$). I am sorry for using the
letter $p$ which already stands for a prime, but this is a standard (and the
$p$ in ``$p_{1},p_{2},p_{3},\ldots$'' always
comes with a subscript).

We let $w_{1},w_{2},w_{3},\ldots\in\Lambda$ be the ``Witt
vector coordinate'' symmetric functions in $\Lambda$; these
are recursively defined by the power-series equality

$\prod_{d\geq1}\left( 1-w_{d}t^{d}\right) ^{-1}=\prod_{i\geq1}\dfrac
{1}{1-x_{i}t}\in\mathbb{Z}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right]
\right] \left[ \left[ t\right] \right] $.

(These $w_{1},w_{2},w_{3},\ldots$ are called $x_{1},x_{2},x_{3},\ldots$ in
§9.63 of Hazewinkel's notes. For a quick overview of their properties,
see Exercise 2.79 in Victor Reiner and YHS, *Hopf Algebras in Combinatorics*, arXiv:1409.8356v3.) Let $\Lambda^{\prime}$ be the subring of $\Lambda$
generated by $w_{p^{0}},w_{p^{1}},w_{p^{2}},\ldots$. Then, the restriction map
$\operatorname*{Ring}\left( \Lambda,\mathbf{k}\right) \rightarrow
\operatorname*{Ring}\left( \Lambda^{\prime},\mathbf{k}\right) $ is
(equivalent to) the canonical quotient map $W\left( \mathbf{k}\right)
\rightarrow W_{p}\left( \mathbf{k}\right) $ from the big Witt vectors
$W\left( \mathbf{k}\right) $ to the $p$-typical Witt vectors $W_{p}\left(
\mathbf{k}\right) $.

It is well-known that $p_{n}=\sum_{d\mid n}dw_{d}^{n/d}$ for every positive
integer $n$. In particular, $p_{p^{k}}=\sum_{i=0}^{k}p^{i}w_{p^{i}}^{p^{k-i}}$
for every $k\in\mathbb{N}$. From this, it is easy to see that for every
$k\in\mathbb{N}$,

**(1)** there exists a positive integer $N$ such that $Nw_{p^{k}}$ is a
polynomial in $p_{p^{0}},p_{p^{1}},\ldots,p_{p^{k}}$ with no constant term.

We now define two ring homomorphisms $\alpha:\Lambda\rightarrow\mathbf{k}$ and
$\beta:\Lambda\rightarrow\mathbf{k}$ by setting

$\alpha\left( f\right) =f\left( 1,0,0,0,\ldots\right) $ and $\beta\left(
f\right) =f\left( -1,0,0,0,\ldots\right) $ for every $f\in\Lambda$

(where the $1$ and the $-1$ are taken in $\mathbf{k}$). (Incidentally,
$\alpha$ is the internal counit of $\Lambda$.)

Now, $\alpha+\beta$ is the ring homomorphism $\Lambda\rightarrow\mathbf{k}$ satisfying

$\left( \alpha+\beta\right) \left( f\right) =f\left( 1,-1,0,0,0,\ldots
\right) $ for every $f\in\Lambda$.

Thus, it is easy to see that $\left( \alpha+\beta\right) \left(
p_{i}\right) =0$ for every odd $i\geq1$. In particular, $\left( \alpha
+\beta\right) \left( p_{p^{i}}\right) =0$ for every $i\in\mathbb{N}$. Thus,
from **(1)** we see that $\left( \alpha+\beta\right) \left( w_{p^{k}
}\right) =0$ for every $k\in\mathbb{N}$ as long as all positive integers are
invertible in $\mathbf{k}$. But we can remove the ``all
positive integers are invertible in $\mathbf{k}$'' condition
(because proving $\left( \alpha+\beta\right) \left( w_{p^{k}}\right) =0$
for $\mathbf{k}=\mathbb{Z}$ is sufficient to ensure that $\left( \alpha
+\beta\right) \left( w_{p^{k}}\right) =0$ holds for every $\mathbf{k}$).
Hence, we obtain $\left( \alpha+\beta\right) \left( w_{p^{k}}\right) =0$
for every $k\in\mathbb{N}$. In other words, the canonical quotient map
$W\left( \mathbf{k}\right) \rightarrow W_{p}\left( \mathbf{k}\right) $
sends $\alpha+\beta$ to the zero of the ring $W_{p}\left( \mathbf{k}\right)
$.

On the other hand, recall that the $\lambda$-ring structure on $W\left(
\mathbf{k}\right) $ comes from plethysm on $\Lambda$. More precisely,

$\left( \lambda^{n}\gamma\right) \left( f\right) =\gamma\left(
e_{n}\left[ f\right] \right) $ for every $n\geq1$, $\gamma\in W\left(
\mathbf{k}\right) $ and $f\in\Lambda$,

where $e_{n}\left[ f\right] $ denotes the plethysm of $f$ into $e_{n}$
(denoted by $e_{n}\circ f$ in Hazewinkel's notes). Applying this to
$n=2$, $\gamma=\alpha+\beta$ and $f=p_{1}$, we obtain

$\left( \lambda^{2}\left( \alpha+\beta\right) \right) \left(
w_{1}\right) =\left( \alpha+\beta\right) \left( \underbrace{e_{2}\left[
w_{1}\right] }_{=e_{2}}\right) =\left( \alpha+\beta\right) \left(
e_{2}\right) $

$=e_{2}\left( 1,-1,0,0,0,\ldots\right) =-1$.

Assume now that $\mathbf{k}\neq0$. Thus, $\left( \lambda^{2}\left(
\alpha+\beta\right) \right) \left( w_{1}\right) =-1\neq0$. Hence, the
canonical quotient map $W\left( \mathbf{k}\right) \rightarrow W_{p}\left(
\mathbf{k}\right) $ sends $\lambda^{2}\left( \alpha+\beta\right) $ to a
nonzero element of the ring $W_{p}\left( \mathbf{k}\right) $.

Thus, the canonical quotient map $W\left( \mathbf{k}\right) \rightarrow
W_{p}\left( \mathbf{k}\right) $ sends $\alpha+\beta$ to zero, but
$\lambda^{2}\left( \alpha+\beta\right) $ to a nonzero element. Therefore,
there exists no $\lambda$-ring structure on $W_{p}\left( \mathbf{k}\right) $
under which this canonical quotient map $W\left( \mathbf{k}\right)
\rightarrow W_{p}\left( \mathbf{k}\right) $ becomes a $\lambda$-ring
morphism. Now, applying this to $\mathbf{k}=\mathbb{F}_{p}$ and recalling that
$W_{p}\left( \mathbb{F}_{p}\right) \cong\mathbb{Z}_{p}$, we see that there
exists no $\lambda$-ring structure on $\mathbb{Z}_{p}$ under which the
canonical quotient map $W\left( \mathbb{F}_{p}\right) \rightarrow
\mathbb{Z}_{p}$ becomes a $\lambda$-ring morphism. As I said, this is all
under the assumption that $p$ is odd, but I would not expect even $p$ to
behave much different.