Yes, regular languages are closed under inverse homomorphism.
Here is a proof.
We start with a DFA of $L$, $X$. To prove that $\varphi^{-1}(L)$ is regular, we will construct a DFA, $Y$ for $\varphi^{-1}(L)$.
$$ X = (Q,\Sigma,\delta_X,q_0,F)$$
So our construction will be as follows:
$$ Y = (Q,\Delta,\delta_Y,q_0,F)$$
Where:
$$\delta_Y(q,w) = \hat{\delta_X}(q,\varphi(w))$$
Or in other words, the transition in Y on w is the result of the sequences of transition that X makes on the string of symbols $\varphi(w)$. So all we need to do now is prove this transition. To do this we will induct on $|w|$. Our hypothesis being $\hat{\delta_Y}(q,w) = \hat{\delta_X}(q,\varphi(w))$.
Base: When $w = \epsilon$:
$$q_0 = \delta_Y(q_0,\epsilon) = \delta_X(q_0,\varphi(\epsilon)) = \delta_X(q_0,\epsilon) = q_0$$
Inductive Step: First let $w = \sigma s$:
$$\hat{\delta_Y}(q_0,w) = \hat{\delta_Y}(\delta_Y(q_0,\sigma),s)$$
Then by our inductive hypothesis:
$$\hat{\delta_Y}(\delta_Y(q_0,\sigma),s) = \hat{\delta_Y}(\delta_X(q_0,\varphi(\sigma)),s)$$
Now by our DFA definition above:
$$\hat{\delta_Y}(\delta_X(q_0,\varphi(\sigma)),s) = \hat{\delta_X}(\delta_X(q_0,\varphi(\sigma)),\varphi(s))$$
Pulling the $\delta$ back in:
$$\hat{\delta_X}(\delta_X(q_0,\varphi(\sigma)),\varphi(s)) = \hat{\delta_X}(q_0,\varphi(\sigma)\varphi(s))$$
Now by the definition of homomorphism:
$$\hat{\delta_X}(q_0,\varphi(\sigma)\varphi(s)) = \hat{\delta_X}(q_0,\varphi(\sigma s)) = \hat{\delta_X}(q_0,\varphi(w))$$