What is the effect of aliasing on the magnitude of the autocorrelation?
Decimating before calculating the autocorrelation, in the presence of noise, is inferior to calculating the autocorrelation using the full dataset. Assume that the signal of interest is embedded in white noise. The vector $x[n], n = 0, 1, ..., N-1$ consists of samples from a discrete random process. The autocorrelation function of the vector $x[n]$ is:

$$
A_x[k] = \frac{1}{N-k} \sum_{i=0}^{N-1-k} x[i]x[i+k]
$$

That is, $k$ is the lag used for the autocorrelation calculation. In your proposed scenario, you are decimating the autocorrelation function output by a factor $D$ (i.e. you are only calculating the function for lags $0, D, 2D, ...$) and comparing that result to the autocorrelation function of $x[n]$ decimated by the same factor $D$. Let $x_d[n]$ be the decimated sequence; its autocorrelation function is:

$$
A_{x_d}[k] = \frac{D}{N-k} \sum_{i=0}^{\frac{N-1-k}{D}} x[iD]x[(i+k)D]
$$

(for simplicity here, I have assumed that $D$ is a factor of $N$ in the above equation)

Your inquiry can be written as:

$$
A_x[kD] \stackrel{?}{\approx} A_{x_d}[k]
$$

$$
\frac{1}{N-kD} \sum_{i=0}^{N-1-kD} x[i]x[i+kD] \stackrel{?}{\approx} \frac{D}{N-k} \sum_{i=0}^{\frac{N-1-k}{D}} x[iD]x[(i+k)D]
$$

Looking at this qualitatively, the summation on the left hand side has more terms than its counterpart on the right side. If $x[n]$ is second-order stationary, then the expected value of each term in each sum is the same; the act of averaging multiple samples that have the same expected value increases the signal to noise ratio. Stated a little differently, you can think the terms in each sum as samples from a new random process:

$$
y[n] = x[n]x[n+kD]
$$

Since the noise present in $x[n]$ is white, the expected value of $y[n]$ is the true autocorrelation of the signal of interest at lag $kD$. Therefore, we would like to accurately estimate the expected value of $y[n]$. Our method for doing so is by calculating a sample average; it can easily be shown that the variance in the sample average estimator decreases given a larger sample size, converging to the actual expected value as the number of samples tends to $\infty$.

So, if there is white noise present in the signal (which is often the case), you're going to get a better estimate of the underlying signal's second-order statistics by using a larger sample size in the calculation (this might sound intuitively obvious). In the context of your two approaches, this is accomplished by using the full, non-decimated signal in the autocorrelation calculation and decimating afterward (i.e. only calculating the result for certain lag values).