Yeah, it's fine. What maSigPro does in this case is to reformulate the data. Your design has columns that are not useful, and maSigPro is smart enough to ignore them. If you run it under the debugger, you can see what happens to your design matrix:

Browse[1]> design
$dis
              Time Time2
Control_3H_1     3     9
Control_3H_2     3     9
Control_3H_3     3     9
Control_9H_1     9    81
Control_9H_2     9    81
Control_9H_3     9    81
Control_27H_1   27   729
Control_27H_2   27   729
Control_27H_3   27   729

$groups.vector
[1] "Group" "Group"

$edesign
              Time Replicate Group
Control_3H_1     3         1     1
Control_3H_2     3         1     1
Control_3H_3     3         1     1
Control_9H_1     9         2     1
Control_9H_2     9         2     1
Control_9H_3     9         2     1
Control_27H_1   27         3     1
Control_27H_2   27         3     1
Control_27H_3   27         3     1

And then for each gene it fits a model using the reformulated data, which if you look at design$dis is fitting a linear and a quadratic term (time + time^2) to account for any possible curvature.

 model.glm <- glm(y ~ ., data = dis, family = family, epsilon = epsilon)

And the fitted model looks like this, where you can see that it's estimating an intercept, a time and a time^ component:

 model.glm

Call:  glm(formula = y ~ ., family = family, data = dis, epsilon = epsilon)

Coefficients:
(Intercept)         Time        Time2  
  -0.456452     0.121108    -0.003772  

Degrees of Freedom: 8 Total (i.e. Null);  6 Residual
Null Deviance:      0.5132 
Residual Deviance: 0.1998   AIC: -0.7258

It then fits a reduced model

 model.glm.0

Call:  glm(formula = y ~ 1, family = family, epsilon = epsilon)

Coefficients:
(Intercept)  
    0.08806  

Degrees of Freedom: 8 Total (i.e. Null);  8 Residual
Null Deviance:      0.5132 
Residual Deviance: 0.5132   AIC: 3.762

That doesn't contain any time coefficients, and then uses the anova function to do a Chi-square test comparing the full and reduced models, which in effect asks if the model that includes time fits the data better than a model with just an intercept (in this case a flat line at 0.088).

 anova(model.glm.0, model.glm, test = "Chisq")
Analysis of Deviance Table

Model 1: y ~ 1
Model 2: y ~ Time + Time2
  Resid. Df Resid. Dev Df Deviance Pr(>Chi)   
1         8    0.51318                        
2         6    0.19985  2  0.31333 0.009064 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

And you can see that this gene is significant before adjusting for multiplicity, but will not be significant after the adjustment.

This test is a likelihood ratio test (LRT), and is not that common for a gaussian fit - people normally use t-tests for conventional linear regression, like this:

 summary(model.glm)

Call:
glm(formula = y ~ ., family = family, data = dis, epsilon = epsilon)

Coefficients:
             Estimate Std. Error t value Pr(>|t|)  
(Intercept) -0.456452   0.194693  -2.344   0.0575 .
Time         0.121108   0.039486   3.067   0.0220 *
Time2       -0.003772   0.001244  -3.033   0.0230 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 0.03330824)

    Null deviance: 0.51318  on 8  degrees of freedom
Residual deviance: 0.19985  on 6  degrees of freedom
AIC: -0.72585

Where you can see that the p-values for both the linear and quadratic term are significant. But maSigPro allows you to use count data as well, in which case you would use either the negative binomial or Poisson distributions, and in which case one could argue that the LRT is a better statistic than the Wald. In addition, using an LRT provides one p-value rather than two, so it's easier to say if a gene is significant or not based on a single p-value.